{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=\\left( m-2 \\right){{x}^{2}}$ n\u1eb1m ph\u00eda tr\u00ean tr\u1ee5c ho\u00e0nh n\u1ebfu","select":["A. $m<2$ ","B. $m>2$","C. $m<-2$ ","D. $m>-2$ "],"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai24/lv1/img\/D24.png' \/><\/center><br\/>\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=ax^2$ n\u1eb1m ph\u00eda tr\u00ean tr\u1ee5c ho\u00e0nh n\u1ebfu $a>0$<br\/>Suy ra \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=\\left( m-2 \\right){{x}^{2}}$ n\u1eb1m ph\u00eda tr\u00ean tr\u1ee5c ho\u00e0nh n\u1ebfu $m-2>0\\Leftrightarrow m>2$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span> <\/span>","column":4}]}],"id_ques":1021},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>H\u00ecnh v\u1ebd th\u1ec3 hi\u1ec7n \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y={{x}^{2}}$ v\u00e0 $y=x+2$ tr\u00ean c\u00f9ng m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 l\u00e0:","select":["A. <img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai24/lv1/img\/D24.2.png' \/> ","B. <img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai24/lv1/img\/D24.1a.png' \/>","C. <img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai24/lv1/img\/D24.3.png' \/> "],"hint":"","explain":"<span class='basic_left'>B\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y=x^2$<br\/><table><tr><td>$x$<\/td><td>$-2$<\/td><td>$-1$<\/td><td>$0$<\/td><td>$1$<\/td><td>$2$<\/td><\/tr><tr><td>$y=x^2$<\/td><td>$4$<\/td><td>$1$<\/td><td>$0$<\/td><td>$1$<\/td><td>$4$<\/td><\/tr><\/tr><\/table><br\/>B\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y=x+2$<br\/><table><tr><td>$x$<\/td><td>$0$<\/td><td>$-2$<\/td><\/tr><tr><td>$y=x+2$<\/td><td>$2$<\/td><td>$0$<\/td><\/tr><\/tr><\/table>Ta c\u00f3 \u0111\u1ed3 th\u1ecb c\u1ee7a hai h\u00e0m s\u1ed1 $y={{x}^{2}}$ v\u00e0 $y=x+2$ l\u00e0:<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai24/lv1/img\/D24.1a.png' \/><\/center><br\/><br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span> <\/span>","column":3}]}],"id_ques":1022},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-1"],["1"],["2"],["4"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>X\u00e1c \u0111\u1ecbnh giao \u0111i\u1ec3m $A,$ $B$ c\u1ee7a parabol $y=x^2$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $y=x+2$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $A$ (_input_;_input_), $B$ (_input_;_input_) v\u1edbi \u0111i\u1ec3m $A$ c\u00f3 ho\u00e0nh \u0111\u1ed9 \u00e2m ","hint":"","explain":"<span class='basic_left'><b>C\u00e1ch 1:<\/b> \u01af\u1edbc l\u01b0\u1ee3ng b\u1eb1ng h\u00ecnh v\u1ebd:<br\/>Theo k\u1ebft qu\u1ea3 c\u00e2u 2, ta c\u00f3 \u0111\u1ed3 th\u1ecb parabol $y=x^2$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $y=x+2$ l\u00e0:<br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai24/lv1/img\/D24.1.png' \/><\/center><br\/>Suy ra giao \u0111i\u1ec3m c\u1ee7a parabol $y=x^2$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $y=x+2$ l\u00e0 $A(-1;1)$ v\u00e0 $B(2;4)$<br\/><b>C\u00e1ch 2: <\/b> X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,{{x}^{2}}=x+2 \\\\ & \\Leftrightarrow {{x}^{2}}-x-2=0 \\\\ \\end{align}$<br\/> V\u00ec $a-b+c=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 $x=-1$ v\u00e0 $x=2$<br\/>Thay $x=-1$ v\u00e0o $y=x+2,$ ta \u0111\u01b0\u1ee3c: $y=1.$ Suy ra giao \u0111i\u1ec3m th\u1ee9 nh\u1ea5t l\u00e0 $A(-1;1)$<br\/>Thay $x=2$ v\u00e0o $y=x+2,$ ta \u0111\u01b0\u1ee3c: $y=4$. Suy ra giao \u0111i\u1ec3m th\u1ee9 nh\u1ea5t l\u00e0 $B(2;4)$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $-1;$ $1;$ $2;$ $4.$<\/span><\/span>"}]}],"id_ques":1023},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["f","f","t"]],"list":[{"point":5,"image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["Ph\u01b0\u01a1ng tr\u00ecnh $2x^2-9=0$ c\u00f3 nghi\u1ec7m k\u00e9p","Ph\u01b0\u01a1ng tr\u00ecnh $x^2-x+2$ c\u00f3 t\u1ed5ng c\u00e1c nghi\u1ec7m b\u1eb1ng $1;$ t\u00edch c\u00e1c nghi\u1ec7m b\u1eb1ng $2;$","Ph\u01b0\u01a1ng tr\u00ecnh $x^2+mx-1=0$ c\u00f3 nghi\u1ec7m v\u1edbi m\u1ecdi $m.$"],"hint":"","explain":["Sai v\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $\\Delta =0+4.2.9=72>0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t","<br\/>Sai v\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $\\Delta =1-4.2=-7<0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m","<br\/>\u0110\u00fang v\u00ec $\\Delta ={{m}^{2}}+4>0$ v\u1edbi m\u1ecdi $m$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 nghi\u1ec7m v\u1edbi m\u1ecdi $m$"]}]}],"id_ques":1024},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["-7"],["10"],["9"],["3"],["5"],["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-7x+10=0$ sau b\u1eb1ng c\u00e1ch \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng:<br\/>$a=$_input_; $b=$_input_; $c=$_input_<br\/>$\\Delta =$_input_. Suy ra $\\sqrt{\\Delta }=$ _input_ <br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: $x_1=$_input_; $x_2=$_input_ bi\u1ebft $x_1>x_2$<\/span>","hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-7x+10=0$ c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 $a=1;b=-7;c=10$ <br\/>$\\Delta ={{\\left( -7 \\right)}^{2}}-4.10=9>0.$ <br\/>Suy ra $\\sqrt{\\Delta }=3$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0:<br\/> ${{x}_{1}}=\\dfrac{7+3}{2}=5;{{x}_{2}}=\\dfrac{7-3}{2}=2$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $1;$ $-7;$ $10;$ $9;$ $3;$ $5;$ $7.$ <\/span><\/span>"}]}],"id_ques":1025},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{2}}+2x-m=0$<br\/>N\u1ed1i t\u1eeb ho\u1eb7c c\u1ee5m t\u1eeb \u1edf c\u1ed9t b\u00ean tr\u00e1i v\u1edbi c\u1ed9t b\u00ean ph\u1ea3i \u0111\u1ec3 \u0111\u01b0\u1ee3c kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang<\/span>","title_trans":"","audio":"","temp":"matching","correct":[["3","1","4","2"]],"list":[{"point":5,"image":"","left":["Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m khi","Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t khi","Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p khi","Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m khi"],"right":["$m>-1$","$m<-1$","$m\\ge -1$ ","$m=-1$"],"top":70,"hint":"X\u00e9t d\u1ea5u $\\Delta '$ ","explain":"<span class='basic_left'>Ta c\u00f3: $\\Delta '=1+m$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m $\\Leftrightarrow \\Delta '\\ge 0\\Leftrightarrow 1+m\\ge 0\\Leftrightarrow m\\ge -1$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $\\Leftrightarrow \\Delta '>0\\Leftrightarrow 1+m>0\\Leftrightarrow m>-1$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p $\\Leftrightarrow \\Delta '=0\\Leftrightarrow 1+m=0\\Leftrightarrow m=-1$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m $\\Leftrightarrow \\Delta '<0\\Leftrightarrow 1+m<0\\Leftrightarrow m<-1$<\/span> "}]}],"id_ques":1026},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $x_{1} = -1; x_{2}= \\dfrac{7}{3}$","B. $x_{1} = 1; x_{2}= -\\dfrac{7}{3}$","C. $x_{1} = 1; x_{2}= \\dfrac{7}{3}$"],"ques":"<span class='basic_left'><br\/>Cho ph\u01b0\u01a1ng tr\u00ecnh: $3x^2-4x+m=0$<br\/><b> C\u00e2u a: <\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi $m=-7$<br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x_1 =$?; $x_2=$?<\/span>","hint":"","explain":"<span class='basic_left'>V\u1edbi $m=-7,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $3{{x}^{2}}-4x-7=0$ <br\/>V\u00ec $a-b+c=3-(-4)-7=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=-1;{{x}_{2}}=\\dfrac{7}{3}$<\/span>"}]}],"id_ques":1027},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh: $3x^2-4x+m=0$<br\/><b>C\u00e2u b:<\/b>T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $x_1;$ $x_2$ th\u1ecfa m\u00e3n $x_1=3x_2$<br\/><b> \u0110\u00e1p s\u1ed1:<\/b> $m=$_input_<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m<br\/>B\u01b0\u1edbc 2: T\u1eeb h\u1ec7 th\u1ee9c Vi-\u00e9t v\u00e0 h\u1ec7 th\u1ee9c \u0111\u00e3 cho, ta t\u00ecm $x_1;x_2$ r\u1ed3i thay v\u00e0o h\u1ec7 th\u1ee9c c\u00f2n l\u1ea1i \u0111\u1ec3 t\u00ecm $m.$<br\/>B\u01b0\u1edbc 3: Ki\u1ec3m tra $m$ c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n kh\u00f4ng r\u1ed3i k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3 $\\Delta '=4-3m$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $x_1;x_2$ $\\Leftrightarrow \\Delta '>0\\Leftrightarrow 4-3m>0\\Leftrightarrow m<\\dfrac{4}{3}$ (*)<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=\\dfrac{4}{3}\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{x}_{1}}{{x}_{2}}=\\dfrac{m}{3}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align} \\right.$ <br\/>Theo gi\u1ea3 thi\u1ebft, ta c\u00f3 $x_1=3x_2.$ Thay v\u00e0o (1), ta c\u00f3: $3{{x}_{2}}+{{x}_{2}}=\\dfrac{4}{3}\\Leftrightarrow 4{{x}_{2}}=\\dfrac{4}{3}\\Leftrightarrow {{x}_{2}}=\\dfrac{1}{3}$ <br\/>Suy ra ${{x}_{1}}=3{{x}_{2}}=3.\\dfrac{1}{3}=1$ <br\/>Thay ${{x}_{1}}=1;{{x}_{2}}=\\dfrac{1}{3}$ v\u00e0o (2), ta c\u00f3 $1.\\dfrac{1}{3}=\\dfrac{m}{3}\\Leftrightarrow m=1$ (th\u1ecfa m\u00e3n (*))<br\/>V\u1eady $m=1$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $x_1;x_2$ th\u1ecfa m\u00e3n $x_1=3x_2$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ <\/span><\/span>"}]}],"id_ques":1028},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\dfrac{3}{2}$","B. $-\\dfrac{3}{2}$","C. $\\dfrac{1}{2}$"],"ques":"<span class='basic_left'>Bi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh $4x^2+2(3-m)x-3m=0$ nh\u1eadn $-1$ l\u00e0m nghi\u1ec7m. Nghi\u1ec7m c\u00f2n l\u1ea1i l\u00e0 ?","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: V\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $-1$ n\u00ean ta thay $x=-1$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 t\u00ecm $m$<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi $m$ v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c \u0111\u1ec3 t\u00ecm nghi\u1ec7m c\u00f2n l\u1ea1i <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho nh\u1eadn $-1$ l\u00e0 nghi\u1ec7m n\u00ean thay $x=-1$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh, ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,4.{{\\left( -1 \\right)}^{2}}+2\\left( 3-m \\right).\\left( -1 \\right)-3m=0 \\\\ & \\Leftrightarrow 4-6+2m-3m=0 \\\\ & \\Leftrightarrow -m-2=0 \\\\ & \\Leftrightarrow m=-2 \\\\ \\end{align}$ <br\/>V\u1edbi $m=-2,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $4{{x}^{2}}+10x+6=0$<br\/>V\u00ec $a-b+c=4-10+6=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $x=-1$ v\u00e0 $x=\\dfrac{-6}{4}=-\\dfrac{3}{2}$ <br\/>V\u1eady nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x=-\\dfrac{3}{2}$<\/span>"}]}],"id_ques":1029},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $2x^4-5x^2+2=0$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0: ","select":["A. $S=\\left\\{ \\sqrt{2};-\\sqrt{2} \\right\\}$ ","B. $S=\\left\\{ \\sqrt{2};-\\sqrt{2};2;-2 \\right\\}$","C. $S=\\left\\{ \\sqrt{2};-\\sqrt{2};\\dfrac{\\sqrt{2}}{2};-\\dfrac{\\sqrt{2}}{2} \\right\\}$","D. $S=\\left\\{ \\dfrac{\\sqrt{2}}{2};-\\dfrac{\\sqrt{2}}{2} \\right\\}$ "],"hint":"\u0110\u00e2y l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng: \u0110\u1eb7t $t=x^2$ \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai \u0111\u1ec3 gi\u1ea3i","explain":"<span class='basic_left'>\u0110\u1eb7t $t={{x}^{2}},\\,t\\ge 0$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{align} & 2{{t}^{2}}-5t+2=0 \\\\ & \\Delta ={{\\left( -5 \\right)}^{2}}-4.2.2=25-16=9 \\\\ & \\Rightarrow \\sqrt{\\Delta }=3 \\\\ & \\Rightarrow {{t}_{1}}=\\dfrac{5+3}{4}=2;{{t}_{2}}=\\dfrac{5-3}{4}=\\dfrac{1}{2} \\\\ \\end{align}$ <br\/>C\u1ea3 hai gi\u00e1 tr\u1ecb \u0111\u1ec1u th\u1ecfa m\u00e3n $t\\ge 0$<br\/> + V\u1edbi $t={{t}_{1}}=2\\Rightarrow {{x}^{2}}=2\\Leftrightarrow x=\\pm \\sqrt{2}$<br\/> + V\u1edbi $t={{t}_{2}}=\\dfrac{1}{2}\\Rightarrow {{x}^{2}}=\\dfrac{1}{2}\\Rightarrow x=\\pm \\dfrac{1}{\\sqrt{2}}$ hay $x=\\pm \\dfrac{\\sqrt{2}}{2}$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ \\sqrt{2};-\\sqrt{2};\\dfrac{\\sqrt{2}}{2};-\\dfrac{\\sqrt{2}}{2} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span><\/span>","column":2}]}],"id_ques":1030},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["2"],["-2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{x}{x+1}+\\dfrac{x}{x-1}=\\dfrac{8}{3}$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_;_input_}<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 2: Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c hai v\u1ebf r\u1ed3i kh\u1eed m\u1eabu th\u1ee9c<br\/>B\u01b0\u1edbc 3: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1eeba nh\u1eadn \u0111\u01b0\u1ee3c<br\/>B\u01b0\u1edbc 4: Trong c\u00e1c gi\u00e1 tr\u1ecb t\u00ecm \u0111\u01b0\u1ee3c c\u1ee7a \u1ea9n, lo\u1ea1i c\u00e1c gi\u00e1 tr\u1ecb kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh. C\u00e1c gi\u00e1 tr\u1ecb th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\pm 1$ <br\/>Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,3x\\left( x-1 \\right)+3x\\left( x+1 \\right)=8\\left( x+1 \\right)\\left( x-1 \\right) \\\\ & \\Leftrightarrow 3{{x}^{2}}-3x+3{{x}^{2}}+3x=8\\left( {{x}^{2}}-1 \\right) \\\\ & \\Leftrightarrow 6{{x}^{2}}=8{{x}^{2}}-8 \\\\ & \\Leftrightarrow 2{{x}^{2}}=8 \\\\ & \\Leftrightarrow {{x}^{2}}=4 \\\\ \\end{align}$<br\/> $\\Leftrightarrow x=\\pm 2$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 2;-2 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2;$ $-2.$ <\/span><\/span>"}]}],"id_ques":1031},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $x_{1} = -2; x_{2}= -\\dfrac{11}{7}$","B. $x_{1} = 2; x_{2}= -\\dfrac{11}{7}$","C. $x_{1} = 1; x_{2}= -\\dfrac{11}{7}$"],"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh:$\\dfrac{2x+5}{4x-3}=\\dfrac{3x+3}{7-x}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 $x_1=$?; $x_2=$?<\/span>","hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\left\\{ 7;\\dfrac{3}{4} \\right\\}$ <br\/>Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\left( 2x+5 \\right)\\left( 7-x \\right)=\\left( 3x+3 \\right)\\left( 4x-3 \\right) \\\\ & \\Leftrightarrow 14x-2{{x}^{2}}+35-5x=12{{x}^{2}}-9x+12x-9 \\\\ & \\Leftrightarrow 14{{x}^{2}}-6x-44=0 \\\\ & \\Leftrightarrow 7{{x}^{2}}-3x-22=0 \\\\ & \\Delta =9+4.7.22=625\\Rightarrow \\sqrt{\\Delta }=25 \\\\ & \\Rightarrow {{x}_{1}}=\\dfrac{3+25}{14}=2;{{x}_{2}}=\\dfrac{3-25}{14}=-\\dfrac{11}{7} \\\\ \\end{align}$ <br\/>C\u1ea3 hai gi\u00e1 tr\u1ecb $x_1; x_2$ \u0111\u1ec1u th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $x$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 2;-\\dfrac{11}{7} \\right\\}$ <\/span>"}]}],"id_ques":1032},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $3{{\\left( {{x}^{2}}+x \\right)}^{2}}-2\\left( {{x}^{2}}+x \\right)-1=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0:<\/span> ","select":["A. $S=\\left\\{ \\dfrac{-1+\\sqrt{5}}{2};\\dfrac{-1-\\sqrt{5}}{2} \\right\\}$","B. $S=\\left\\{ \\dfrac{1+\\sqrt{5}}{2};\\dfrac{1-\\sqrt{5}}{2} \\right\\}$","C. $S=\\left\\{ -1+\\sqrt{5};-1-\\sqrt{5} \\right\\}$","D. $S=\\left\\{ 1+\\sqrt{5};1-\\sqrt{5} \\right\\}$"],"hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 $t={{x}^{2}}+x$ ","explain":"<span class='basic_left'>\u0110\u1eb7t $t={{x}^{2}}+x$, ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $3{{t}^{2}}-2t-1=0$ <br\/>V\u00ec $a+b+c=3-2-1=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $t=1;$ $t=-\\dfrac{1}{3}$ <br\/>+ V\u1edbi $t=1,$ ta c\u00f3 ${{x}^{2}}+x=1\\Leftrightarrow {{x}^{2}}+x-1=0$ <br\/>$\\Delta =5>0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}}=\\dfrac{-1+\\sqrt{5}}{2};{{x}_{2}}=\\dfrac{-1-\\sqrt{5}}{2}$ <br\/>+ V\u1edbi $t=-\\dfrac{1}{3}$ , ta c\u00f3 ${{x}^{2}}+x=-\\dfrac{1}{3}\\Leftrightarrow {{x}^{2}}+x+\\dfrac{1}{3}=0\\Leftrightarrow 3{{x}^{2}}+3x+1=0$ <br\/>$\\Delta =-3<0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ \\dfrac{-1+\\sqrt{5}}{2};\\dfrac{-1-\\sqrt{5}}{2} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A <\/span><\/span>","column":2}]}],"id_ques":1033},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{1}{2}$","B. $-\\dfrac{1}{3}$","C. $\\dfrac{1}{4}$"],"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $2{{x}^{3}}-{{x}^{2}}+2x-1=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{$?$\\}$<\/span>","hint":"Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i ph\u01b0\u01a1ng tr\u00ecnh th\u00e0nh nh\u00e2n t\u1eed \u0111\u1ec3 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{aligned} & \\,\\,\\,\\,\\,\\,2{{x}^{3}}-{{x}^{2}}+2x-1=0 \\\\ & \\Leftrightarrow \\left( 2{{x}^{3}}-{{x}^{2}} \\right)+\\left( 2x-1 \\right)=0 \\\\ & \\Leftrightarrow {{x}^{2}}\\left( 2x-1 \\right)+\\left( 2x-1 \\right)=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+1 \\right)\\left( 2x-1 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}+1=0 \\\\ & 2x-1=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}=-1\\,\\left( \\text{v\u00f4 nghi\u1ec7m v\u00ec}\\,x^2 \\ge 0\\,\\text{v\u00e0}\\, -1 < 0\\right) \\\\ & x=\\dfrac{1}{2} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ \\dfrac{1}{2} \\right\\}$<\/span>"}]}],"id_ques":1034},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","2"]],"list":[{"point":5,"img":"","ques":"T\u00ecm hai s\u1ed1 $a$ v\u00e0 $b$ bi\u1ebft $a+b=9;$ $a.b=14$ ","hint":"","column":2,"number_true":2,"select":["A. $a=7;$ $b=2$","B. $a=2;$ $b=7$","C. $a=-2;$ $b=-7$","D. $a=-7;$ $b=-2$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u00e0i to\u00e1n: T\u00ecm hai s\u1ed1 bi\u1ebft t\u1ed5ng $S$ v\u00e0 t\u00edch $P$<br\/>Hai s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai $X^2-SX+P=0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Hai s\u1ed1 $a, b$ c\u1ea7n t\u00ecm l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai $X^2-9X+14=0$<br\/>$\\Delta =81-4.14=25$<br\/> Suy ra ${{X}_{1}}=\\dfrac{9+5}{2}=7;{{X}_{2}}=\\dfrac{9-5}{2}=2$ <br\/>V\u1eady $a=7;b=2$ ho\u1eb7c $a=2;b=7$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A v\u00e0 B <\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t: <\/span>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 t\u1ed3n t\u1ea1i hai s\u1ed1 $a, b$ l\u00e0 ${{S}^{2}}\\ge 4P$ <\/span>"}]}],"id_ques":1035},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'>N\u1ed1i m\u1ed7i \u00f4 \u1edf c\u1ed9t tr\u00e1i v\u1edbi m\u1ed9t \u00f4 \u1edf c\u1ed9t ph\u1ea3i \u0111\u1ec3 \u0111\u01b0\u1ee3c c\u00e1c kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang.<br\/>Cho ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai $ax^2+bx+c=0$ (1) v\u1edbi $a\\ne 0$. \u0110\u1eb7t $S=-\\dfrac{b}{a};P=\\dfrac{c}{a}$ <\/span>","title_trans":"","audio":"","temp":"matching","correct":[["3","4","2","1"]],"list":[{"point":5,"image":"","left":["Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u khi","Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t c\u00f9ng d\u1ea5u khi","Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t khi","Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t khi"],"right":["$\\Delta >0;P>0;S<0$","$\\Delta >0;P>0;S>0$","$P<0$","$\\Delta >0$ v\u00e0 $P>0$ "],"top":100,"hint":"","explain":"<span class='basic_left'>+ Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u $P<0$<br\/>+ Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t c\u00f9ng d\u1ea5u $\\Delta >0$ v\u00e0 $P>0$ <br\/>+ Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t l\u00e0 $\\Delta >0;P>0;S>0$<br\/>+ Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t l\u00e0 $\\Delta >0;P>0;S<0$<\/span> "}]}],"id_ques":1036},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-2mx+{{\\left( m-1 \\right)}^{2}}=0$ c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng.<\/span>","select":["A. $m\\le \\dfrac{1}{2}$ ","B. $m\\ge \\dfrac{1}{2};m\\ne 1$ ","C. $m\\le -\\dfrac{1}{2}$ ","D. $m\\ge -\\dfrac{1}{2};m\\ne 1$ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/> Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t l\u00e0 $\\Delta >0;P>0;S>0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/>$\\Delta '={{m}^{2}}-{{\\left( m-1 \\right)}^{2}}={{m}^{2}}-\\left( {{m}^{2}}-2m+1 \\right)=2m-1$ <br\/>$S=-\\dfrac{b}{a}=2m;P=\\dfrac{c}{a}={{\\left( m-1 \\right)}^{2}}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng<br\/> $\\Leftrightarrow \\left\\{ \\begin{aligned} & \\Delta '\\ge 0 \\\\ & S>0 \\\\ & P>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2m-1\\ge 0 \\\\ & 2m>0 \\\\ & {{\\left( m-1 \\right)}^{2}}>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ge \\dfrac{1}{2} \\\\ & m>0 \\\\ & m\\ne 1 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ge \\dfrac{1}{2} \\\\ & m\\ne 1 \\\\ \\end{aligned} \\right.$<br\/> V\u1eady v\u1edbi $m\\ge \\dfrac{1}{2};m\\ne 1$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1037},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["8"],["6"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>T\u00ecm hai s\u1ed1 bi\u1ebft t\u1ed5ng c\u1ee7a ch\u00fang l\u00e0 $14$ v\u00e0 t\u1ed5ng b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a ch\u00fang l\u00e0 $100.$<br\/>Hai s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 _input_v\u00e0 _input_<\/span>","hint":"Gi\u1ea3i b\u00e0i to\u00e1n b\u1eb1ng c\u00e1ch l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi \u1ea9n l\u00e0 m\u1ed9t trong hai s\u1ed1 c\u1ea7n t\u00ecm","explain":"<span class='basic_left'>G\u1ecdi s\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 $x$ th\u00ec s\u1ed1 th\u1ee9 hai l\u00e0 $14-x$<br\/>V\u00ec t\u1ed5ng b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a hai s\u1ed1 b\u1eb1ng $100$ n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{align} & {{x}^{2}}+{{\\left( 14-x \\right)}^{2}}=100 \\\\ & \\Leftrightarrow {{x}^{2}}+196-28x+{{x}^{2}}=100 \\\\ & \\Leftrightarrow 2{{x}^{2}}-28x+96=0 \\\\ & \\Leftrightarrow {{x}^{2}}-14x+48=0 \\\\ \\end{align}$<br\/> $\\Delta '={{7}^{2}}-48=1$<br\/> Suy ra ${{x}_{1}}=7+1=8;{{x}_{2}}=7-1=6$<br\/>V\u1eady hai s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 $8$ v\u00e0 $6$<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $8;$ $6.$<\/span><\/span>"}]}],"id_ques":1038},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"],["12"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng ch\u00e9o l\u00e0 $13$ $m$ v\u00e0 chi\u1ec1u d\u00e0i h\u01a1n chi\u1ec1u r\u1ed9ng $7$ $m.$ T\u00ednh chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a m\u1ea3nh \u0111\u1ea5t.<br\/><b> \u0110\u00e1p s\u1ed1:<\/b> Chi\u1ec1u r\u1ed9ng: _input_ $(m);$ chi\u1ec1u d\u00e0i: _input_ $(m)$<\/span>","hint":"","explain":"<span class='basic_left'>G\u1ecdi chi\u1ec1u r\u1ed9ng c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 $x$ $(m),$ $x>0$<br\/>V\u00ec chi\u1ec1u d\u00e0i h\u01a1n chi\u1ec1u r\u1ed9ng l\u00e0 $7$ $m$ n\u00ean chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 $x+7$ $(m)$<br\/>V\u00ec \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 c\u1ea1nh huy\u1ec1n c\u1ee7a tam gi\u00e1c vu\u00f4ng ch\u1ee9a \u0111\u01b0\u1eddng ch\u00e9p \u0111\u00f3 n\u00ean \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago, ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>${{x}^{2}}+{{\\left( x+7 \\right)}^{2}}=169$<br\/> $\\Leftrightarrow {{x}^{2}}+{{x}^{2}}+14x+49=169$<br\/>$ \\Leftrightarrow 2{{x}^{2}}+14x-120=0$<br\/>$ \\Leftrightarrow {{x}^{2}}+7x-60=0$<br\/>$ \\Delta =49+4.60=289\\Rightarrow \\sqrt{\\Delta }=17 $<br\/>$\\Rightarrow {{x}_{1}}=\\dfrac{-7+17}{2}=5;{{x}_{2}}=\\dfrac{-7-17}{2}=-12 $<br\/>V\u00ec $x>0$ n\u00ean $x=x_1=5$<br\/>V\u1eady chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 $5$ $(m)$ v\u00e0 chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 $x+7=5+7=12$ $(m)$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $5$ v\u00e0 $12.$<\/span><\/span>"}]}],"id_ques":1039},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["12"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>M\u1ed9t ng\u01b0\u1eddi \u0111i xe \u0111\u1ea1p t\u1eeb A \u0111\u1ebfn B c\u00e1ch nhau $24$ $km.$ Khi \u0111i t\u1eeb B tr\u1edf v\u1ec1 A, ng\u01b0\u1eddi \u0111\u00f3 t\u0103ng v\u1eadn t\u1ed1c th\u00eam $4$ $km\/h$ so v\u1edbi l\u00fac \u0111i, v\u00ec v\u1eady th\u1eddi gian v\u1ec1 \u00edt h\u01a1n th\u1eddi gian \u0111i l\u00e0 $30$ ph\u00fat. T\u00ednh v\u1eadn t\u1ed1c xe \u0111\u1ea1p khi \u0111i t\u1eeb A \u0111\u1ebfn B<br\/><b>\u0110\u00e1p s\u1ed1:<\/b>_input_ $(km\/h)$<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u1ea3ng ph\u00e2n t\u00edch chuy\u1ec3n \u0111\u1ed9ng: <br\/> <table><tr><th>C\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng<br><\/th><th>V\u1eadn t\u1ed1c $(km\/h)$<br><\/th><th>Qu\u00e3ng \u0111\u01b0\u1eddng $(km)$<br><\/th><th>Th\u1eddi gian (gi\u1edd)<br><\/th><\/tr><tr><th>\u0110i t\u1eeb A \u0111\u1ebfn B<br><\/th><td>$x$<\/td><td>$24$<\/td><td>$\\dfrac{24}{x}$<\/td><\/tr><tr><th>\u0110i t\u1eeb B v\u1ec1 A<br><\/th><td>$x+4$<\/td><td>$24$<\/td><td>$\\dfrac{24}{x+4}$<\/td><\/tr><\/table><br\/>L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb th\u1eddi gian l\u00fac \u0111i v\u00e0 l\u00fac v\u1ec1 c\u1ee7a xe \u0111\u1ea1p.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi v\u1eadn t\u1ed1c c\u1ee7a xe \u0111\u1ea1p khi \u0111i t\u1eeb A \u0111\u1ebfn B l\u00e0 $x$ $(km\/h),$ $x>0$<br\/>V\u00ec khi \u0111i t\u1eeb B v\u1ec1 A, ng\u01b0\u1eddi \u0111\u00f3 t\u0103ng v\u1eadn t\u1ed1c th\u00eam $4$ $km\/h$ so v\u1edbi l\u00fac \u0111i n\u00ean v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi \u0111i xe \u0111\u1ea1p khi t\u1eeb B v\u1ec1 A l\u00e0 $x+4$ $(km\/h)$<br\/>V\u00ec qu\u00e3ng \u0111\u01b0\u1eddng AB d\u00e0i $24km$ n\u00ean th\u1eddi gian khi \u0111i t\u1eeb A \u0111\u1ebfn B l\u00e0 $\\dfrac{24}{x}$ (gi\u1edd) v\u00e0 th\u1eddi gian khi \u0111i t\u1eeb B v\u1ec1 A l\u00e0 $\\dfrac{24}{x+4}$ (gi\u1edd)<br\/>Do th\u1eddi gian v\u1ec1 \u00edt h\u01a1n th\u1eddi gian \u0111i l\u00e0 $30$ ph\u00fat $=\\dfrac{1}{2}$ gi\u1edd n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\dfrac{24}{x}-\\dfrac{24}{x+4}=\\dfrac{1}{2} \\\\ & \\Rightarrow 2.24.\\left( x+4 \\right)-2.24.x=x\\left( x+4 \\right) \\\\ & \\Leftrightarrow 48x+192-48x={{x}^{2}}+4x \\\\ & \\Leftrightarrow {{x}^{2}}+4x-192=0 \\\\ & \\Delta '=196 \\\\ & \\Rightarrow {{x}_{1}}=-2+14=12;{{x}_{2}}=-2-14=-16 \\\\ \\end{align}$<br\/> V\u00ec $x>0$ n\u00ean $x=x_1=12$<br\/>V\u1eady v\u1eadn t\u1ed1c c\u1ee7a xe \u0111\u1ea1p khi \u0111i t\u1eeb A \u0111\u1ebfn B l\u00e0 $12$ $(km\/h)$<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $12.$<\/span><\/span>"}]}],"id_ques":1040}],"lesson":{"save":0,"level":1}}