{"segment":[{"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":[[["4"],["-8"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\left( x-1 \\right)\\left( x+5 \\right)\\left( x-3 \\right)\\left( x+7 \\right)=297$ <br\/><b>\u0110\u00e1p s\u1ed1: <\/b> 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\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc 4 d\u1ea1ng $\\left( x+a \\right)\\left( x+b \\right)\\left( x+c \\right)\\left( x+d \\right)=m$ v\u1edbi $a+b=c+d$<br\/><b>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i:<\/b><br\/>B\u01b0\u1edbc 1: Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1ee3c vi\u1ebft th\u00e0nh: $\\left[ {{x}^{2}}+\\left( a+b \\right)x+ab \\right]\\left[ {{x}^{2}}+\\left( c+d \\right)x+cd \\right]=m.$<br\/> \u0110\u1eb7t $t={{x}^{2}}+\\left( a+b \\right)x,$ ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: $\\left( t+ab \\right)\\left( t+cd \\right)=m$ <br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm $t,$ t\u1eeb \u0111\u00f3 t\u00ecm $x$ b\u1eb1ng c\u00e1ch gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+\\left( a+b \\right)x=t$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\begin{aligned} & \\left( x-1 \\right)\\left( x+5 \\right)\\left( x-3 \\right)\\left( x+7 \\right)=297 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+4x-5 \\right)\\left( {{x}^{2}}+4x-21 \\right)=297 \\\\ \\end{aligned}$<br\/>\u0110\u1eb7t $t={{x}^{2}}+4x$, ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\left( t-5 \\right)\\left( t-21 \\right)=297\\Leftrightarrow {{t}^{2}}-26t-192=0$<br\/>$\\Delta '={{13}^{2}}+192=361\\Rightarrow \\sqrt{\\Delta '}=19$ <br\/>Suy ra $\\left[ \\begin{aligned} & t=13+19=32 \\\\ & t=13-19=-6 \\\\ \\end{aligned} \\right.$ <br\/>V\u1edbi $t=32,$ ta c\u00f3 ${{x}^{2}}+4x-32=0\\Leftrightarrow \\left[ \\begin{aligned} & x=4 \\\\ & x=-8 \\\\ \\end{aligned} \\right.$ <br\/>V\u1edbi $t=-6,$ ta c\u00f3 ${{x}^{2}}+4x+6=0$ , $\\Delta '=-2<0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ 4;-8 \\right\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4$ v\u00e0 $-8$<\/span><\/span>"}]}],"id_ques":961},{"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":[[["-4"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{\\left( x+3 \\right)}^{4}}+{{\\left( x+5 \\right)}^{4}}=2$ <br\/><b> \u0110\u00e1p s\u1ed1:<\/b> Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $ x =$_input_<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 d\u1ea1ng ${{\\left( x+a \\right)}^{4}}+{{\\left( x+b \\right)}^{4}}=c$ <br\/><b>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i:<\/b><br\/>\u0110\u1eb7t $t=x+\\dfrac{a+b}{2},$ suy ra $x=t-\\dfrac{a+b}{2}$<br\/> Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh: ${{\\left( t+\\dfrac{a-b}{2} \\right)}^{4}}+{{\\left( t-\\dfrac{a-b}{2} \\right)}^{4}}=c$<br\/> Khai tri\u1ec3n v\u00e0 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng \u0111\u1ed1i v\u1edbi $t.$<br\/>\u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: ${{\\left( x+ y \\right)}^{4}}={{x}^{4}}+ 4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}+ 4x{{y}^{3}}+{{y}^{4}}$ \u0111\u1ec3 khai tri\u1ec3n ph\u01b0\u01a1ng tr\u00ecnh.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110\u1eb7t $t=x+\\dfrac{3+5}{2}=x+4\\Rightarrow x=t-4$ <br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{align} & {{\\left( t-1 \\right)}^{4}}+{{\\left( t+1 \\right)}^{4}}=2 \\\\ & \\Leftrightarrow \\left( {{t}^{4}}-4{{t}^{3}}+6{{t}^{2}}-4t+1 \\right)+\\left( {{t}^{4}}+4{{t}^{3}}+6{{t}^{2}}+4t+1 \\right)=2 \\\\ & \\Leftrightarrow 2{{t}^{4}}+12{{t}^{2}}=0 \\\\ & \\Leftrightarrow {{t}^{2}}\\left( {{t}^{2}}+6 \\right)=0 \\\\ & \\Leftrightarrow t=0 \\,(\\text{do}\\, t^2 + 6 > 0) \\\\ & \\Leftrightarrow x+4=0 \\\\ & \\Leftrightarrow x=-4 \\\\ \\end{align}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $-4$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-4.$<\/span><\/span>"}]}],"id_ques":962},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{4}}+24x+32=0$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{ 2;-4;1+\\sqrt{5};1-\\sqrt{5} \\right\\}$","B. $S=\\left\\{ 2;-4;-1+\\sqrt{5};-1-\\sqrt{5} \\right\\}$","C. $S=\\left\\{ -1+\\sqrt{5};-1-\\sqrt{5} \\right\\}$","D. $S=\\left\\{ 1+\\sqrt{5};1-\\sqrt{5} \\right\\}$"],"hint":"\u0110\u01b0a hai v\u1ebf v\u1ec1 l\u0169y th\u1eeba c\u00f9ng b\u1eadc","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{4}}=24x+32 \\\\ & \\Leftrightarrow {{x}^{4}}+4{{x}^{2}}+4=4{{x}^{2}}+24x+36 \\\\ & \\Leftrightarrow {{\\left( {{x}^{2}}+2 \\right)}^{2}}={{\\left( 2x+6 \\right)}^{2}} \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}+2=2x+6\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{x}^{2}}+2=-2x-6\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>$\\begin{aligned} & \\left( 1 \\right)\\Leftrightarrow {{x}^{2}}-2x-4=0 \\\\ & \\Delta '=5\\Rightarrow x=1\\pm \\sqrt{5} \\\\ \\end{aligned}$ <br\/>$\\begin{aligned} & \\left( 2 \\right)\\Leftrightarrow {{x}^{2}}+2x+8=0 \\\\ & \\Delta '=-7<0 \\\\ \\end{aligned}$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh (2) v\u00f4 nghi\u1ec7m.<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ 1+\\sqrt{5};1-\\sqrt{5} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span><\/span>","column":2}]}],"id_ques":963},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{4}}+6{{x}^{3}}+7{{x}^{2}}+6x+1=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{ \\dfrac{5\\pm \\sqrt{21}}{2} \\right\\}$","B. $S=\\left\\{ \\dfrac{-5\\pm \\sqrt{21}}{2} \\right\\}$","C. $S=\\left\\{ -5\\pm \\sqrt{21} \\right\\}$","D. $S=\\left\\{ 5\\pm \\sqrt{21} \\right\\}$"],"hint":"Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed1i x\u1ee9ng b\u1eadc 4. Ta chia c\u1ea3 hai v\u1ebf cho $x^2$ kh\u00e1c $0,$ r\u1ed3i \u0111\u1eb7t \u1ea9n ph\u1ee5 $t=x+\\dfrac{1}{x}$ \u0111\u1ec3 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed1i x\u1ee9ng d\u1ea1ng: $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+ bx+a=0,\\left( a\\ne 0 \\right)$ <br\/><b>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i:<\/b><br\/>V\u00ec $x =0$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh n\u00ean chia c\u1ea3 hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh cho $x^2,$ ta \u0111\u01b0\u1ee3c:<br\/>$a\\left( {{x}^{2}}+\\dfrac{1}{{{x}^{2}}} \\right)+b\\left( x+ \\dfrac{1}{x} \\right)+c=0$ <br\/>\u0110\u1eb7t $t=x+ \\dfrac{1}{x}\\Rightarrow {{t}^{2}}={{x}^{2}}+\\dfrac{1}{{{x}^{2}}}+ 2\\Rightarrow {{x}^{2}}+\\dfrac{1}{{{x}^{2}}}={{t}^{2}}- 2.$ <br\/>Khi \u0111\u00f3 ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai: $a\\left( {{t}^{2}}-2 \\right)+bt+c=0$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>X\u00e9t $x=0$, ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh $1=0$ (v\u00f4 l\u00ed). Suy ra $x =0$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh <br\/>Chia c\u1ea3 hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh cho $x^2,$ ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & \\Leftrightarrow {{x}^{2}}+6x+7+\\dfrac{6}{x}+\\dfrac{1}{{{x}^{2}}}=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+\\dfrac{1}{{{x}^{2}}} \\right)+6\\left( x+\\dfrac{1}{x} \\right)+7=0 \\\\ \\end{aligned}$<br\/>\u0110\u1eb7t $t=x+\\dfrac{1}{x}\\Rightarrow {{t}^{2}}={{x}^{2}}+\\dfrac{1}{{{x}^{2}}}+2\\Rightarrow {{x}^{2}}+\\dfrac{1}{{{x}^{2}}}={{t}^{2}}-2$ <br\/>Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai:<br\/>$\\begin{aligned} & {{t}^{2}}-2+6t+7=0 \\\\ & \\Leftrightarrow {{t}^{2}}+6t+5=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t=-1 \\\\ & t=-5 \\\\ \\end{aligned} \\right.\\,(\\text{do}a-b+c=0) \\\\ \\end{aligned}$ <br\/>+ V\u1edbi $t=-1,$ ta c\u00f3 $x+\\dfrac{1}{x}=-1\\Leftrightarrow {{x}^{2}}+x+1=0$ (ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m)<br\/>+ V\u1edbi $t=-5,$ ta c\u00f3 $x+\\dfrac{1}{x}=-5\\Leftrightarrow {{x}^{2}}+5x+1=0\\Leftrightarrow x=\\dfrac{-5\\pm \\sqrt{21}}{2}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ \\dfrac{-5\\pm \\sqrt{21}}{2} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>N\u1ebfu $\\alpha$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed1i x\u1ee9ng th\u00ec $\\dfrac{1}{\\alpha}$ c\u0169ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<\/span>","column":2}]}],"id_ques":964},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{4}}-4{{x}^{2}}+x+2=0\\,\\,\\,\\,\\left( 1 \\right)$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{ 1;2;\\dfrac{1+\\sqrt{5}}{2};\\dfrac{1-\\sqrt{5}}{2} \\right\\}$ ","B. $S=\\left\\{ 1;-2;\\dfrac{-1+\\sqrt{5}}{2};\\dfrac{-1-\\sqrt{5}}{2} \\right\\}$ ","C. $S=\\left\\{ 1;-2;\\dfrac{1+\\sqrt{5}}{2};\\dfrac{1-\\sqrt{5}}{2} \\right\\}$ ","D. $S=\\left\\{ -1;2;\\dfrac{1+\\sqrt{5}}{2};\\dfrac{1-\\sqrt{5}}{2} \\right\\}$ "],"hint":"Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{4}}-4{{x}^{2}}+x+2={{x}^{2}}\\left( {{x}^{2}}-4 \\right)+\\left( x+2 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{2}}\\left( x+2 \\right)\\left( x-2 \\right)+\\left( x+2 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( x+2 \\right)\\left( {{x}^{3}}-2{{x}^{2}}+1 \\right) \\\\ \\end{aligned}$<br\/> Ph\u00e2n t\u00edch ti\u1ebfp ${{x}^{3}}-2{{x}^{2}}+1$ th\u00e0nh nh\u00e2n t\u1eed \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & {{x}^{3}}-2{{x}^{2}}+1={{x}^{3}}-x^2-{{x}^{2}}+1 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{2}}\\left( x-1 \\right)-\\left( x-1 \\right)\\left( x+1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( x-1 \\right)\\left( {{x}^{2}}-x-1 \\right) \\\\ \\end{aligned}$ <br\/>Do \u0111\u00f3: <br\/>$\\begin{aligned} & \\left( 1 \\right)\\Leftrightarrow \\left( x+2 \\right)\\left( x-1 \\right)\\left( {{x}^{2}}-x-1 \\right)=0 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow \\left[ \\begin{aligned} & x+2=0 \\\\ & x-1=0 \\\\ & {{x}^{2}}-x-1=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=-2 \\\\ & x=1 \\\\ & x=\\dfrac{1\\pm \\sqrt{5}}{2} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 $S=\\left\\{ 1;-2;\\dfrac{1+\\sqrt{5}}{2};\\dfrac{1-\\sqrt{5}}{2} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>C\u00e1ch gi\u1ea3i kh\u00e1c c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1): Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i ph\u01b0\u01a1ng tr\u00ecnh th\u00e0nh m\u1ed9t t\u00edch c\u1ee7a hai \u0111a th\u1ee9c b\u1eadc hai:<br\/>$\\begin{aligned} & {{x}^{4}}-4{{x}^{2}}+x+2=\\left( {{x}^{4}}-4{{x}^{2}}+4 \\right)-{{x}^{2}}+\\left( {{x}^{2}}+x-2 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{\\left( {{x}^{2}}-2 \\right)}^{2}}-{{x}^{2}}+\\left( {{x}^{2}}+x-2 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( {{x}^{2}}-2-x \\right)\\left( {{x}^{2}}-2+x \\right)+\\left( {{x}^{2}}+x-2 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( {{x}^{2}}-2+x \\right)\\left[ \\left( {{x}^{2}}-2-x \\right)+1 \\right] \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( {{x}^{2}}+x-2 \\right)\\left( {{x}^{2}}-x-1 \\right) \\\\ \\end{aligned}$<\/span>","column":2}]}],"id_ques":965},{"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":[[["0"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\sqrt{x+4}+\\sqrt{x+1}=\\sqrt{2x+9}$ <br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x=$_input_<\/span> ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i:<br\/>B\u01b0\u1edbc 1: \u0110\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ngh\u0129a<br\/>B\u01b0\u1edbc 2: B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $\\left\\{ \\begin{aligned} & x+4\\ge 0 \\\\ & x+1\\ge 0 \\\\ & 2x+9\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge -4 \\\\ & x\\ge -1 \\\\ & x\\ge -\\dfrac{9}{2} \\\\ \\end{aligned} \\right.\\Leftrightarrow x\\ge -1$ <br\/>B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & x+4+x+1+2\\sqrt{x+4}.\\sqrt{x+1}=2x+9 \\\\ & \\Leftrightarrow 2x+5+2\\sqrt{\\left( x+4 \\right)\\left( x+1 \\right)}=2x+9 \\\\ & \\Leftrightarrow \\sqrt{{{x}^{2}}+5x+4}=2 \\\\ & \\Leftrightarrow {{x}^{2}}+5x+4=4 \\\\ & \\Leftrightarrow {{x}^{2}}+5x=0 \\\\ & \\Leftrightarrow x\\left( x+5 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0\\,\\,\\,\\,\\,\\,\\,\\left( \\text{th\u1ecfa m\u00e3n}\\, \\right) \\\\ & x=-5\\,\\,\\,\\,\\,\\left( \\text{lo\u1ea1i}\\, \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m l\u00e0 $x=0$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0.$ <\/span><\/span>"}]}],"id_ques":966},{"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":[[["6"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\sqrt[3]{x+2}-\\sqrt[3]{7-x}=1$ <br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x=$_input_<\/span> ","hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 $\\sqrt[3]{x+2}=a,\\,\\,\\sqrt[3]{7-x}=b$. Khi \u0111\u00f3 t\u00ednh $a^3+b^3$ v\u00e0 $a-b.$ T\u1eeb \u0111\u00f3 t\u00ecm $x$","explain":"<span class='basic_left'>\u0110\u1eb7t $\\sqrt[3]{x+2}=a,\\,\\,\\sqrt[3]{7-x}=b$. Ta c\u00f3:<br\/>${{a}^{3}}+{{b}^{3}}=\\left( x+2 \\right)+\\left( 7-x \\right)=9\\,$ v\u00e0 $a-b=1$<br\/>Thay $a=b+1$ v\u00e0o ${{a}^{3}}+{{b}^{3}}=9,$ ta c\u00f3: <br\/>$\\begin{aligned} & {{\\left( b+1 \\right)}^{3}}+{{b}^{3}}=9 \\\\ & \\Leftrightarrow 2{{b}^{3}}+3{{b}^{2}}+3b-8=0 \\\\ & \\Leftrightarrow \\left( b-1 \\right)\\left( 2{{b}^{2}}+5b+8 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & b-1=0 \\\\ & 2{{b}^{2}}+5b+8=0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>+ V\u1edbi $b-1=0$ th\u00ec $b=1$ <br\/> Suy ra $\\sqrt[3]{7-x}=1\\Leftrightarrow 7-x=1\\Leftrightarrow x=6$<br\/> + V\u1edbi $2{{b}^{2}}+5b+8=0:$ V\u00ec $\\Delta =-39<0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m l\u00e0 $x=6$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $6.$ <\/span><\/span>"}]}],"id_ques":967},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n d\u01b0\u1edbi d\u1ea1ng s\u1ed1 nguy\u00ean ho\u1eb7c s\u1ed1 th\u1eadp ph\u00e2n","temp":"fill_the_blank_random","correct":[[["3"],["-1,2"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{9}{{{x}^{2}}}+\\dfrac{9}{{{\\left( x+2 \\right)}^{2}}}=10$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{$_input_;_input_$\\}$<\/span> ","hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 $t=x+1$ ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\left\\{ 0;-2 \\right\\}$ <br\/>\u0110\u1eb7t $t=x+1.$ Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\dfrac{9}{{{\\left( t-1 \\right)}^{2}}}+\\dfrac{9}{{{\\left( t+1 \\right)}^{2}}}=10$ <br\/>$\\begin{aligned} & \\Leftrightarrow 9{{\\left( t+1 \\right)}^{2}}+9{{\\left( t-1 \\right)}^{2}}=10{{\\left( t+1 \\right)}^{2}}{{\\left( t-1 \\right)}^{2}} \\\\ & \\Leftrightarrow 9\\left( {{t}^{2}}+2t+1 \\right)+9\\left( {{t}^{2}}-2t+1 \\right)=10{{\\left( {{t}^{2}}-1 \\right)}^{2}} \\\\ & \\Leftrightarrow 18{{t}^{2}}+18=10\\left( {{t}^{4}}-2{{t}^{2}}+1 \\right) \\\\ & \\Leftrightarrow 10{{t}^{4}}-38{{t}^{2}}-8=0 \\\\ & \\Leftrightarrow 5{{t}^{2}}-19t-4=0 \\\\ \\end{aligned}$ <br\/>$\\Delta =441\\Rightarrow \\sqrt{\\Delta }=21$<br\/> Suy ra ${{t}_{1}}=\\dfrac{19+21}{10}=4;{{t}_{2}}=\\dfrac{19-21}{10}=-\\dfrac{1}{5}$<br\/> + V\u1edbi $t={{t}_{1}}=4$ th\u00ec $x+1=4\\Leftrightarrow x=3$ (th\u1ecfa m\u00e3n)<br\/>+ V\u1edbi $t={{t}_{2}}=-\\dfrac{1}{5}$ th\u00ec $x+1=-\\dfrac{1}{5}\\Leftrightarrow x=-\\dfrac{6}{5}=-1,2$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m l\u00e0 $S=\\left\\{ 3;-1,2 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3;$ $-1,2.$<\/span><\/span>"}]}],"id_ques":968},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $x_{1} = 3; x_{2}= -\\dfrac{36}{13}$","B. $x_{1} = 1; x_{2}= -\\dfrac{36}{13}$","C. $x_{1} = 2; x_{2}= -\\dfrac{36}{13}$"],"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{3}{3+\\sqrt{9-{{x}^{2}}}}-\\dfrac{2}{3-\\sqrt{9-{{x}^{2}}}}=\\dfrac{1}{x}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 $x_1=$?;$x_2=$?<\/span> ","hint":"Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $-3\\le x\\le 3,x\\ne 0$ <br\/>Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & \\dfrac{3\\left( 3-\\sqrt{9-{{x}^{2}}} \\right)}{{{x}^{2}}}-\\dfrac{2\\left( 3+\\sqrt{9-{{x}^{2}}} \\right)}{{{x}^{2}}}=\\dfrac{1}{x} \\\\ & \\Leftrightarrow \\dfrac{3-5\\sqrt{9-{{x}^{2}}}}{{{x}^{2}}}=\\dfrac{x}{{{x}^{2}}} \\\\ & \\Leftrightarrow 3-5\\sqrt{9-{{x}^{2}}}=x \\\\ & \\Leftrightarrow 5\\sqrt{9-{{x}^{2}}}=3-x \\\\ \\end{aligned}$ <br\/>V\u1edbi \u0111i\u1ec1u ki\u1ec7n $x\\le 3,$ b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\,25\\left( 9-{{x}^{2}} \\right)={{\\left( 3-x \\right)}^{2}} \\\\ & \\Leftrightarrow 225-25{{x}^{2}}=9-6x+{{x}^{2}} \\\\ & \\Leftrightarrow 26{{x}^{2}}-6x-216=0 \\\\ & \\Leftrightarrow 13{{x}^{2}}-3x-108=0 \\\\ \\end{aligned}$<br\/>$\\Delta =5625\\Rightarrow \\sqrt{\\Delta }=75$ <br\/>Suy ra ${{x}_{1}}=\\dfrac{3+75}{26}=3$; ${{x}_{2}}=\\dfrac{3-75}{26}=-\\dfrac{36}{13}$ <br\/>C\u1ea3 hai nghi\u1ec7m \u0111\u1ec1u th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n.<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ 3;\\dfrac{-36}{13} \\right\\}$ <\/span>"}]}],"id_ques":969},{"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":[[["3"],["3"],["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"T\u00ecm $x,y,z$ th\u1ecfa m\u00e3n c\u1ea3 hai ph\u01b0\u01a1ng tr\u00ecnh:<br\/> $\\begin{aligned} & x+y+z=9; \\\\ & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=27 \\\\ \\end{aligned}$ <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $x=$_input_; $y=$_input_; $z=$_input_ ","hint":"B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 nh\u1ea5t v\u00e0 nh\u00e2n hai v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 hai v\u1edbi $3.$ T\u1eeb \u0111\u00f3 k\u1ebft h\u1ee3p l\u1ea1i \u0111\u1ec3 t\u00ecm $x,y,z$","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$\\begin{aligned} & +)\\,x+y+z=9 \\\\ & \\Leftrightarrow {{\\left( x+y+z \\right)}^{2}}=81 \\\\ & \\Leftrightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\\left( xy+yz+zx \\right)=81\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ \\end{aligned}$ <br\/>$\\begin{aligned} & +)\\, {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=27 \\\\ & \\Leftrightarrow 3{{x}^{2}}+3{{y}^{2}}+3{{z}^{2}}=81\\,\\,\\left( 2 \\right) \\\\ \\end{aligned}$ <br\/>T\u1eeb (1) v\u00e0 (2), ta c\u00f3:<br\/> $\\begin{aligned} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\\left( xy+yz+zx \\right)=3{{x}^{2}}+3{{y}^{2}}+3{{z}^{2}} \\\\ & \\Leftrightarrow 2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-2xy-2yz-2zx=0 \\\\ & \\Leftrightarrow {{\\left( x-y \\right)}^{2}}+{{\\left( y-z \\right)}^{2}}+{{\\left( z-x \\right)}^{2}}=0 \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & x-y=0 \\\\ & y-z=0 \\\\ & z-x=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow x=y=z \\\\ \\end{aligned}$ <br\/>K\u1ebft h\u1ee3p v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh $x+y+z=9, $ ta suy ra $x=y=z=3$<br\/>Th\u1eed l\u1ea1i: Thay $x=y=z$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=27$ th\u00ec th\u1ecfa m\u00e3n<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3;3;3.$<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>C\u00e1ch 2:<br\/>$\\begin{aligned} & x+y+z=9 \\\\ & \\Leftrightarrow 6x+6y+6z=54\\,\\,\\,\\left( 3 \\right) \\\\ \\end{aligned}$<br\/>Tr\u1eeb t\u1eebng v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=27$ cho t\u1eebng v\u1ebf c\u1ee7a (3), ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x-6y-6z+27=0 \\\\ & \\Leftrightarrow {{\\left( x-3 \\right)}^{2}}+{{\\left( y-3 \\right)}^{2}}+{{\\left( z-3 \\right)}^{2}}=0 \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & x-3=0 \\\\ & y-3=0 \\\\ & z-3=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow x=y=z=3 \\\\ \\end{aligned}$<br\/> Th\u1eed l\u1ea1i $x=y=z=3$ th\u1ecfa m\u00e3n.<\/span>"}]}],"id_ques":970}],"lesson":{"save":0,"level":3}}