{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-8"],["3"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/9.jpg' \/><\/center>Cho h\u00e0m s\u1ed1 $y=x^2-6x+1$. T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00f3.<br\/>\u0110\u00e1p \u00e1n: $y_{min}=$ _input_ khi $x =$ _input_","hint":"\u0110\u01b0a h\u00e0m s\u1ed1 \u0111\u00e3 cho v\u1ec1 d\u1ea1ng $y = a + [f(x)]^2$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: D\u1ef1a v\u00e0o h\u1eb1ng \u0111\u1eb3ng th\u1ee9c ${{A}^{2}}\\pm 2AB+{{B}^{2}}={{\\left( A\\pm B \\right)}^{2}}$ \u0111\u1ec3 \u0111\u01b0a bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng $T=a+{{\\left[ f\\left( x \\right) \\right]}^{2}}$<br\/>B\u01b0\u1edbc 2: V\u00ec $f{{\\left[ \\left( x \\right) \\right]}^{2}}\\ge 0\\,\\,\\forall \\,x$ n\u00ean $T\\ge a$ . Suy ra ${{T}_{\\min }}=a\\Leftrightarrow f\\left( x \\right)=0$. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/>$\\begin{align}\\,\\,y&={{x}^{2}}-6x+1\\\\&=(x^2-6x+9)-9+1\\\\&={{\\left( x-3 \\right)}^{2}}-8\\ge -8\\\\ \\end{align}$ <br\/>${{y}_{\\min }}=-8\\Leftrightarrow {{\\left( x-3 \\right)}^{2}}=0\\Leftrightarrow x=3$ <br\/>V\u1eady ${{y}_{\\min }}=-8\\Leftrightarrow x=3$<br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-8$ v\u00e0 $3$.<\/span>"}]}],"id_ques":21},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho h\u00e0m s\u1ed1 $y=\\dfrac{{{x}^{2}}-3x+2}{\\sqrt{x-2}}$ . T\u00ecm $x$ \u0111\u1ec3 $y=0$. ","select":["A. $x= 1$","B. $x = 2$","C. $x= 1$ ho\u1eb7c $x = 2$","D. $x\\in \\varnothing$"],"hint":"$\\dfrac{a}{b}=0\\Leftrightarrow a=0$","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 c\u1ee7a $x$ \u0111\u1ec3 h\u00e0m s\u1ed1 c\u00f3 ngh\u0129a.<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh. <br\/>B\u01b0\u1edbc 3: K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 t\u00ecm ra nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 $x \u2013 2 > 0 \\Leftrightarrow x > 2$<br\/>$\\begin{aligned} y=0 & \\Leftrightarrow \\dfrac{{{x}^{2}}-3x+2}{\\sqrt{x-2}}=0 \\\\ & \\Leftrightarrow \\dfrac{{{x}^{2}}-2x-x+2}{\\sqrt{x-2}}=0 \\\\ & \\Rightarrow x^2-2x-x+2=0 \\\\ & \\Leftrightarrow x(x-2)-(x-2)=0 \\\\ & \\Leftrightarrow \\left( x-1 \\right)\\left( x-2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-1=0 \\\\ & x-2=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=1\\,\\,\\,\\,\\,(lo\u1ea1i) \\\\ & x=2\\,\\,\\,\\,\\,(lo\u1ea1i) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Suy ra kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $x$ \u0111\u1ec3 $y= 0$. <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":2}]}],"id_ques":22},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"T\u00ecm $m$ \u0111\u1ec3 h\u00e0m s\u1ed1 $y = (m^2-5 )x+2$ ngh\u1ecbch bi\u1ebfn tr\u00ean $\\mathbb{R}$","select":["A. $m > \\sqrt{5}$ ho\u1eb7c $m < -\\sqrt{5}$","B. $m > \\sqrt{5}$","C. $m < -\\sqrt{5}$","D. $-\\sqrt{5}\\,<\\,m \\,<\\,\\sqrt{5}$"],"hint":": Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $a < 0$. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t $ y= ax +b\\, (a\\ne 0)$ ngh\u1ecbch bi\u1ebfn khi $a < 0$<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2 -b^2= (a-b)(a+b)$ <br\/>B\u01b0\u1edbc 3: Chia tr\u01b0\u1eddng h\u1ee3p gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>H\u00e0m s\u1ed1 $y = (m^2-5 )x+2$ ngh\u1ecbch bi\u1ebfn tr\u00ean $\\mathbb{R}$ <br\/> $\\Leftrightarrow {{m}^{2}-5}<0\\Leftrightarrow \\left( m-\\sqrt{5} \\right)\\left( m+\\sqrt{5} \\right) < 0$ <br\/>Tr\u01b0\u1eddng h\u1ee3p 1: $\\left\\{ \\begin{aligned} & m-\\sqrt{5}>0 \\\\ & m+\\sqrt{5}<0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m>\\sqrt{5} \\\\ & m< -\\sqrt{5} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\,$ (lo\u1ea1i) <br\/>Tr\u01b0\u1eddng h\u1ee3p 2: $\\left\\{ \\begin{aligned} & m-\\sqrt{5} < 0 \\\\ & m+\\sqrt{5}> 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m < \\sqrt{5} \\\\ & m >- \\sqrt{5}\\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow -\\sqrt{5} < m <\\sqrt{5} $ <br\/>Do \u0111\u00f3 v\u1edbi $-\\sqrt{5}\\,<\\,m \\,<\\,\\sqrt{5}$ th\u00ec h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean $\\mathbb{R}$. <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":2}]}],"id_ques":23},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/4.jpg' \/><\/center> Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$ cho \u0111i\u1ec3m $A (3; 4)$. \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n $OA$ l\u00e0 _input_","hint":"H\u1ea1 $AH \\bot Ox$ t\u1ea1i $H$. <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Py - ta - go v\u00e0o tam gi\u00e1c vu\u00f4ng $OAH$. ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/D921_TB5.png' \/><\/center><br\/>H\u1ea1 $AH \\bot Ox$ t\u1ea1i $H$ $\\Rightarrow H (3; 0)$.<br\/>Theo h\u00ecnh v\u1ebd ta c\u00f3 $AH= 4; OH =3$.<br\/>X\u00e9t $\\Delta AHO$ vu\u00f4ng t\u1ea1i $H$ c\u00f3:<br\/> $\\begin{align} AO^2&=AH^2+OH^2\\,\\,(\u0110\u1ecbnh\\,\\,l\u00fd\\,\\,Py - ta - go)\\\\&=16+9\\\\&=25 \\\\ \\Rightarrow&AO= 5\\\\ \\end{align}$ <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $5$.<\/span>"}]}],"id_ques":24},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["3"],["-2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'> Cho $A (-3; 2), B (1; 4)$. $ABCD$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh nh\u1eadn g\u1ed1c $O$ l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng. <br\/><b>C\u00e2u 1:<\/b> T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $C$ l\u00e0 (_input_; _input_) <\/span>","hint":"V\u1ebd h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9, l\u1ea5y \u0111i\u1ec3m $A$ v\u00e0 $B$, l\u1ea5y \u0111i\u1ec3m $C$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $A$ qua $O$.","explain":"<span class='basic_left'><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/D921_TB3.png' \/><\/center><\/center><br\/>L\u1ea5y \u0111i\u1ec3m $A(-3; 2)$ v\u00e0 \u0111i\u1ec3m $B (1; 4)$ tr\u00ean h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 $Oxy$ nh\u01b0 h\u00ecnh v\u1ebd.<br\/>V\u00ec $ABCD$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh c\u00f3 t\u00e2m \u0111\u1ed1i x\u1ee9ng l\u00e0 $O(0;0)$ n\u00ean $C$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $A$ qua $O$ c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0 $(3;-2)$<br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $3$ v\u00e0 $-2$.<\/span><br\/>C\u00e1ch 2: Ngo\u00e0i ra, ta c\u00f3 th\u1ec3 t\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $C$ b\u1eb1ng c\u00e1ch \u00e1p d\u1ee5ng c\u00f4ng th\u1ee9c t\u1ecda \u0111\u1ed9 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng.<br\/>V\u00ec $A$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $C$ qua $O$, suy ra $O$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$. <br\/>T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $C$ l\u00e0 :<br\/>$\\left\\{ \\begin{aligned} & {{x}_{C}}=2{{x}_{O}}-{{x}_{A}} \\\\ & {{y}_{C}}=2{{y}_{O}}-{{y}_{A}} \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{C}}=2.0-\\left( -3 \\right) \\\\ & {{y}_{C}}=2.0-2 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{C}}=3 \\\\ & {{y}_{C}}=-2 \\\\ \\end{aligned} \\right.$ <br\/>$\\Rightarrow C\\left( 3;-2 \\right)$ <br\/><span class='basic_green'>Ghi nh\u1edb:<\/span> $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$. T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $M$ \u0111\u01b0\u1ee3c cho b\u1edfi:<br\/>$ \\left\\{ \\begin{aligned} & 2{{x}_{M}}={{x}_{A}}+{{x}_{B}} \\\\ & 2{{y}_{M}}={{y}_{A}}+{{y}_{B}} \\\\ \\end{aligned} \\right.$ <\/span>"}]}],"id_ques":25},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["-1"],["-4"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'> Cho $A (-3; 2), B (1; 4)$. $ABCD$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh nh\u1eadn g\u1ed1c $O$ l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng. <br\/><b>C\u00e2u 2:<\/b>T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $D$ l\u00e0 (_input_; _input_) <\/span>","hint":"V\u1ebd h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9, l\u1ea5y \u0111i\u1ec3m $A$ v\u00e0 $B$, l\u1ea5y \u0111i\u1ec3m $D$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $B$ qua $O$.","explain":"<span class='basic_left'>Ta c\u00f3 h\u00ecnh v\u1ebd sau: <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/D921_TB3.png' \/><\/center><br\/>L\u1ea5y \u0111i\u1ec3m $A (-3; 2)$ v\u00e0 \u0111i\u1ec3m $B (1; 4)$ tr\u00ean h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 $Oxy$ nh\u01b0 h\u00ecnh v\u1ebd.<br\/>V\u00ec $ABCD$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh c\u00f3 t\u00e2m \u0111\u1ed1i x\u1ee9ng l\u00e0 $O(0;0)$ n\u00ean $D$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $B$ qua $O$ c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0 $(-1;-4)$<br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-1$ v\u00e0 $-4$.<\/span><br\/><span class='basic_left'>Nh\u1eadn x\u00e9t:<\/span>T\u01b0\u01a1ng t\u1ef1 c\u00e2u tr\u00ean, ta c\u00f3 th\u1ec3 t\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m D b\u1eb1ng c\u00e1ch:<br\/>V\u00ec $B$ \u0111\u1ed1i x\u1ee9ng v\u1edbi $D$ qua $O$ $\\Rightarrow O$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BD$. <br\/>T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $D$ l\u00e0 :<br\/>$\\left\\{ \\begin{aligned} & {{x}_{D}}=2{{x}_{O}}-{{x}_{B}} \\\\ & {{y}_{D}}=2{{y}_{O}}-{{y}_{B}} \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{D}}=2.0-1 \\\\ & {{y}_{D}}=2.0-4 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{D}}=-1 \\\\ & {{y}_{D}}=-4 \\\\ \\end{aligned} \\right.$ <br\/>$\\Rightarrow D\\left( -1;-4 \\right)$ <\/span> <\/span>"}]}],"id_ques":26},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\sqrt{13}$","B. $2\\sqrt{13}$","C. $3\\sqrt{13}$","D. $4\\sqrt{13}$"],"ques":"<span class='basic_left'> Cho $A (-3; 2), B (1; 4)$. $ABCD$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh nh\u1eadn g\u1ed1c $O$ l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng. <br\/><b>C\u00e2u 3:<\/b> \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n $AC$ l\u00e0 bao nhi\u00eau?<\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>H\u1ea1 $AH \\bot Oy$ t\u1ea1i $H$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Py - ta - go v\u00e0o tam gi\u00e1c $OAH$ vu\u00f4ng t\u1ea1i $H$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3 h\u00ecnh v\u1ebd sau: <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/D921_TB4.png' \/><\/center><br\/>H\u1ea1 $AH \\bot Oy$ t\u1ea1i $H$ $\\Rightarrow H (0; 2)$<br\/>Theo h\u00ecnh v\u1ebd ta c\u00f3 $AH= 3; OH =2$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Py - ta - go v\u00e0o tam gi\u00e1c $OAH$ vu\u00f4ng t\u1ea1i $H$, ta c\u00f3: $AO^2=AH^2+OH^2$ <br\/> $\\Rightarrow AO^2=9+4=13$<br\/>$\\Rightarrow AO=\\sqrt{13}$ <br\/> $ \\Rightarrow AC=2AO = 2\\sqrt{13}$<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 $2\\sqrt{13}$.<\/span>"}]}],"id_ques":27},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $y_{max} = \\dfrac{13}{4}, x = \\dfrac{3}{2}$","B. $y_{max} = \\dfrac{11}{4}, x = \\dfrac{1}{2}$","C. $y_{max} = \\dfrac{13}{5}, x = \\dfrac{3}{2}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/9.jpg' \/><\/center>Cho h\u00e0m s\u1ed1 $y=-x^2+3x+1$. T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00f3. <br\/>\u0110\u00e1p \u00e1n: $y_{max}=$ ? khi $x$ = ?","hint":"\u0110\u01b0a h\u00e0m s\u1ed1 \u0111\u00e3 cho v\u1ec1 d\u1ea1ng $y = b\u2212[f(x)]^2 $","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: D\u1ef1a v\u00e0o h\u1eb1ng \u0111\u1eb3ng th\u1ee9c ${{A}^{2}}\\pm 2AB+{{B}^{2}}={{\\left( A\\pm B \\right)}^{2}}$ \u0111\u1ec3 \u0111\u01b0a bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng $T=b-{{\\left[ f\\left( x \\right) \\right]}^{2}}$ <br\/>B\u01b0\u1edbc 2: V\u00ec $-f{{\\left[ \\left( x \\right) \\right]}^{2}}\\le 0\\,\\,\\forall x$ n\u00ean $T\\le b$ . <br\/> Suy ra ${{T}_{\\text{max}}}=b\\Leftrightarrow f\\left( x \\right)=0$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3 $y=-{{x}^{2}}+3x+1\\,$$=-(x^2-2.\\dfrac{3}{2}.x+\\dfrac{9}{4})+\\dfrac{9}{4}+1\\,$$=-{{\\left( x-\\dfrac{3}{2} \\right)}^{2}}+\\dfrac{13}{4}\\le \\dfrac{13}{4}$ <br\/>${{y}_{\\max }}=\\dfrac{13}{4}\\Leftrightarrow {{\\left( x -\\dfrac{3}{2} \\right)}^{2}}=0\\,$$\\Leftrightarrow x=\\dfrac{3}{2}$ <br\/>V\u1eady ${{y}_{\\max }}=\\dfrac{13}{4}\\Leftrightarrow x= \\dfrac{3}{2}$<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 y_{max} = \\dfrac{13}{4} khi x = \\dfrac{3}{2}$<\/span><\/span>"}]}],"id_ques":28},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\dfrac{1}{3}$","B. $\\dfrac{2}{3}$","C. $\\dfrac{4}{3}$"],"ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $f(x)= 2x+1$ v\u00e0 $g(x) =x + 3$<br\/>T\u00ecm $a$ sao cho $f(a+1) =g(2-a)$.<br\/>Gi\u00e1 tr\u1ecb c\u1ee7a $a$ t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 ? <\/span>","hint":"T\u00ednh $f(a+1)$ v\u00e0 $g(2-a)$ v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $f(a+1)=g(2\u2212a)$ \u0111\u1ec3 t\u00ecm a. ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & f\\left( a+1 \\right)=2(a +1)+1=2a+3 \\\\ & g\\left( 2-a \\right)=2-a+3=5-a\\\\ &\\text{Ta c\u00f3:}\\\\&\\,\\,\\,\\,\\,\\, f\\left( a+1 \\right)=g\\left( 2-a \\right)\\\\&\\Leftrightarrow 2a +3 =5-a \\\\ &\\Leftrightarrow 3a=2 \\\\ & \\Leftrightarrow a=\\dfrac{2}{3} \\\\ \\end{align}$<br\/><span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0: $\\dfrac{2}{3}$. <\/span>"}]}],"id_ques":29},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["2"],["2"],["4"],["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho h\u00ecnh v\u1ebd sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/D921_TB2.png' \/><\/center> <span class='basic_left'> Bi\u1ebft \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c $Ox$ v\u00e0 c\u1eaft tr\u1ee5c $Oy$ t\u1ea1i \u0111i\u1ec3m c\u00f3 tung \u0111\u1ed9 $y= 2$ l\u1ea7n l\u01b0\u1ee3t c\u1eaft c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $y=x$ v\u00e0 $y=\\dfrac{x}{2}$ t\u1ea1i hai \u0111i\u1ec3m $A$ v\u00e0 $B$.<br\/><b> C\u00e2u 1: <\/b> T\u1ecda \u0111\u1ed9 $A$ l\u00e0 (_input_;_input_) <br\/> T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $B$ l\u00e0 (_input_; _input_)","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>Thay $y=2$ v\u00e0o t\u1eebng ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u00e3 cho \u0111\u1ec3 t\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a $A$ v\u00e0 $B$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>+ T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $A$. <br\/> Thay $y= 2$ v\u00e0o $y=x$ ta \u0111\u01b0\u1ee3c: $x =2 $ <br\/>Suy ra $A (2 ; 2)$<br\/>+ T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $B$.<br\/> Thay $y = 2$ v\u00e0o $y= \\dfrac{x}{2}$ ta \u0111\u01b0\u1ee3c: $ x= 4$<br\/>Suy ra $B (4; 2)$<br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $2;2;4;2$.<\/span><\/span>"}]}],"id_ques":30},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank_random","correct":[[["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho h\u00ecnh v\u1ebd sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/D921_TB2.png' \/><\/center><span class='basic_left'> Bi\u1ebft \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c $Ox$ v\u00e0 c\u1eaft tr\u1ee5c $Oy$ t\u1ea1i \u0111i\u1ec3m c\u00f3 tung \u0111\u1ed9 $y= 2$ l\u1ea7n l\u01b0\u1ee3t c\u1eaft c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $y=x$ v\u00e0 $y=2x$ t\u1ea1i hai \u0111i\u1ec3m $A$ v\u00e0 $B$.<br\/><b> C\u00e2u 2: <\/b> Di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ l\u00e0 _input_ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","hint":"$S=\\dfrac{1}{2}OH.AB$","explain":"<span class='basic_left'>Theo c\u00e2u 1 tr\u00ean, ta c\u00f3 t\u1ecda \u0111\u1ed9 c\u1ee7a $A$ v\u00e0 $B$ l\u00e0: <br\/>$A(2; 2) , B(4;2 ) \\Rightarrow AB =2$. <br\/>V\u00ec $H(0;2)$ n\u00ean ta c\u00f3 $OH=2$. <br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ l\u00e0: <br\/> $S_{OAB}=\\dfrac {1}{2}.OH.AB=\\dfrac{1}{2}.2.2=2$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch). <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2$.<\/span><\/span>"}]}],"id_ques":31},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["9,3"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho h\u00ecnh v\u1ebd <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/D921_TB2.png' \/><\/center><span class='basic_left'> Bi\u1ebft \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c $Ox$ v\u00e0 c\u1eaft tr\u1ee5c $Oy$ t\u1ea1i \u0111i\u1ec3m c\u00f3 tung \u0111\u1ed9 $y= 2$ l\u1ea7n l\u01b0\u1ee3t c\u1eaft c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $y=x$ v\u00e0 $y=2x$ t\u1ea1i hai \u0111i\u1ec3m $A$ v\u00e0 $B$.<br\/><b> C\u00e2u 3: <\/b> Chu vi tam gi\u00e1c $OAB$ $\\approx$ _input_<br\/> (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t)","hint":"+ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Py - ta - go \u0111\u1ec3 t\u00ednh c\u00e1c c\u1ea1nh $OA, OB$.<br\/>+ Chu vi tam gi\u00e1c $OAB$ b\u1eb1ng $OA +OB+AB$.","explain":"<span class='basic_left'>Theo h\u00ecnh v\u1ebd, $ AH=2, BH=4$<br\/>Theo c\u00e2u 1 tr\u00ean, ta c\u00f3 $A(2; 2) , B(4; 2), OH=2, AB= 2$.<br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago v\u00e0o tam gi\u00e1c $OAH$ v\u00e0 $OBH$, ta c\u00f3: <br\/> $OA =\\sqrt{AH^2+OH^2}=\\sqrt{2^2+2^2}=\\sqrt{8}$<br\/> $OB=\\sqrt{BH^2+OH^2}=\\sqrt{4^2+2^2}=\\sqrt{20}$<br\/>Chu vi tam gi\u00e1c $OAB$ l\u00e0: <br\/> $2 + \\sqrt{8}+ \\sqrt{20}\\approx 9,3$ <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $9,3$.<\/span><\/span>"}]}],"id_ques":32},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"T\u00ecm $m$ \u0111\u1ec3 h\u00e0m s\u1ed1 $y = (m^2 - 4)x+2$ \u0111\u1ed3ng bi\u1ebfn tr\u00ean $\\mathbb{R}$.","select":["A. $m > 2$ ho\u1eb7c $m < -2$ ","B. $m > 2$","C. $m < -2$","D. $-2 < m < 2$"],"hint":"H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t $y= ax +b\\, (a\\ne 0)$ \u0111\u1ed3ng bi\u1ebfn khi $a > 0$. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t $y= ax +b\\, (a\\ne 0)$ \u0111\u1ed3ng bi\u1ebfn khi $a > 0$.<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $a>0$: Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eeb r\u1ed3i chia tr\u01b0\u1eddng h\u1ee3p \u0111\u1ec3 gi\u1ea3i.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>H\u00e0m s\u1ed1 $y = (m^2 - 4)x+2$ \u0111\u1ed3ng bi\u1ebfn tr\u00ean $\\mathbb{R}$ <br\/> $\\Leftrightarrow {{m}^{2}}-4>0\\Leftrightarrow \\left( m-2 \\right)\\left( m+2 \\right)>0$ <br\/>Tr\u01b0\u1eddng h\u1ee3p 1: $\\left\\{ \\begin{aligned} & m-2>0 \\\\ & m+2>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m>2 \\\\ & m>-2 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow m>2$ <br\/>Tr\u01b0\u1eddng h\u1ee3p 2: $\\left\\{ \\begin{aligned} & m-2<0 \\\\ & m+2<0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m<2 \\\\ & m<-2 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow m<-2$<br\/>Do \u0111\u00f3 v\u1edbi $m > 2$ ho\u1eb7c $m < - 2$ th\u00ec h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean $\\mathbb{R}$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":33},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"],["-1"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/9.jpg' \/><\/center>Cho h\u00e0m s\u1ed1 $y=x^2+2x+3$.<br\/>T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00f3.<br\/>\u0110\u00e1p \u00e1n: $y_{min}=$ _input_ khi $x =$ _input_","hint":"\u0110\u01b0a h\u00e0m s\u1ed1 \u0111\u00e3 cho v\u1ec1 d\u1ea1ng $y = a+[f(x)]^2 $.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: D\u1ef1a v\u00e0o h\u1eb1ng \u0111\u1eb3ng th\u1ee9c ${{A}^{2}}\\pm 2AB+{{B}^{2}}={{\\left( A\\pm B \\right)}^{2}}$ \u0111\u1ec3 \u0111\u01b0a bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng $T=a+{{\\left[ f\\left( x \\right) \\right]}^{2}}$<br\/>B\u01b0\u1edbc 2: V\u00ec $f{{\\left[ \\left( x \\right) \\right]}^{2}}\\ge 0\\,\\,\\forall x$ n\u00ean $T\\ge a$ . Suy ra ${{T}_{\\min }}=a\\Leftrightarrow f\\left( x \\right)=0$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3 $y={{x}^{2}}+2x+3={{\\left( x+1 \\right)}^{2}}+2\\ge 2$ <br\/>${{y}_{\\min }}=2\\Leftrightarrow {{\\left( x+1 \\right)}^{2}}=0\\Leftrightarrow x=-1$ <br\/>V\u1eady ${{y}_{\\min }}=2\\Leftrightarrow x=-1$<br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2$ v\u00e0 $-1$.<\/span>"}]}],"id_ques":34},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho h\u00e0m s\u1ed1 $y=-x^2-2x+3$. Gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 t\u1ea1i $x= \\sqrt{3}-1$ l\u00e0 _input_","hint":"Thay $x=\\sqrt{3}-1$ v\u00e0o h\u00e0m s\u1ed1 $y=-x^2-2x+3$.","explain":"<span class='basic_left'> Thay $x=\\sqrt{3}-1$ v\u00e0o h\u00e0m s\u1ed1 $y=-x^2-2x+3$ ta \u0111\u01b0\u1ee3c:<br\/> $\\begin{align} y & =-{{\\left( \\sqrt{3}-1 \\right)}}-2\\left( \\sqrt{3}-1 \\right)+3 \\\\ & =-{{\\left( 3-2\\sqrt{3}+1 \\right)}^{2}}-2\\left( \\sqrt{3}-1 \\right)+3 \\\\ & =-\\left( 4-2\\sqrt{3} \\right)-2\\sqrt{3}+2+3 \\\\ & =-4+2\\sqrt{3}-2\\sqrt{3}+5 \\\\ & =1 \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$.<\/span>"}]}],"id_ques":35},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Khi $x<0$, h\u00e0m s\u1ed1 $y=f(x)= 2x^2$ ngh\u1ecbch bi\u1ebfn. ","select":["A. \u0110\u00fang","B. Sai"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>- L\u1ea5y $x_{1} < x_2$ ($x_1,x_2$ th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh). X\u00e9t hi\u1ec7u $f(x_{1}) -f(x_{2})$ <br\/>+ N\u1ebfu $f(x_{1}) -f(x_{2}) < 0$ th\u00ec $f(x_{1}) < f(x_{2})$: H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn trong kho\u1ea3ng $(a;b)$ <br\/>+ N\u1ebfu $f(x_{1}) -f(x_{2}) > 0$ th\u00ec $f(x_{1}) > f(x_{2})$: H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn trong kho\u1ea3ng $(a;b)$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>V\u1edbi ${{x}_{1}}<{{x}_{2}}<0$, ta c\u00f3: <br\/>$f\\left( {{x}_{1}} \\right)-f\\left( {{x}_{2}} \\right)=2x_{1}^{2}-2x_{2}^{2}\\,$$=2\\left( {{x}_{1}}-{{x}_{2}} \\right)\\left( {{x}_{1}}+{{x}_{2}} \\right)$ <br\/>V\u00ec ${{x}_{1}}<{{x}_{2}}<0$ suy ra :<br\/>$\\left. \\begin{aligned} & {{x}_{1}}+{{x}_{2}}<0 \\\\ & {{x}_{1}}-{{x}_{2}}<0 \\\\ \\end{aligned} \\right\\}\\Rightarrow f\\left( {{x}_{1}} \\right)-f\\left( {{x}_{2}} \\right)>0\\,$$\\Rightarrow f\\left( {{x}_{1}} \\right)>f\\left( {{x}_{2}} \\right)$ <br\/>Do \u0111\u00f3, khi $x<0$ h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn <br\/> <span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang. <\/span><\/span>","column":2}]}],"id_ques":36},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{3}$","B. $2\\sqrt{3}$","C. $3\\sqrt{3}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/9.jpg' \/><\/center>H\u00e0m s\u1ed1 $y=\\sqrt{3}x -mx + 1$ l\u00e0 h\u00e0m h\u1eb1ng khi $m=$ ?","hint":"H\u00e0m s\u1ed1 $y = ax+ b$ l\u00e0 h\u00e0m h\u1eb1ng khi $a= 0$. ","explain":"<span class='basic_left'> Ta c\u00f3: $y=\\sqrt{3}x-mx+1\\,$$\\Leftrightarrow y=x\\left( \\sqrt{3}-m \\right)+1$<br\/>H\u00e0m s\u1ed1 $y=\\sqrt{3}x-mx+1$ l\u00e0 h\u00e0m h\u1eb1ng $\\Leftrightarrow \\sqrt{3}-m=0 \\Leftrightarrow m= \\sqrt{3}$<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $\\sqrt{3}$.<\/span>"}]}],"id_ques":37},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"H\u00e0m s\u1ed1 $y= (m^2-m-2)x+2$ l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t khi:","select":["A. $m \\ne -1$ ","B. $m \\ne 2$","C. $m \\ne -1$ ho\u1eb7c $m \\ne 2$","D. $m \\ne -1$ v\u00e0 $m \\ne 2$"],"hint":"H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t c\u00f3 d\u1ea1ng $y= ax +b\\, (a\\ne 0)$ ","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 h\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0 h\u00e0m b\u1eadc nh\u1ea5t.<br\/>B\u01b0\u1edbc 2: Ph\u00e2n t\u00edch h\u1ec7 s\u1ed1 $a$ th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i t\u00ecm $m$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> H\u00e0m s\u1ed1 $y= (m^2-m-2)x+2$ l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t khi: <br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\,\\,{{m}^{2}}-m-2\\ne 0 \\\\ & \\Leftrightarrow {{m}^{2}}+m-2m-2\\ne 0 \\\\ & \\Leftrightarrow m(m+1)-2(m+1) \\ne 0 \\\\ & \\Leftrightarrow \\left( m+1 \\right)\\left( m-2 \\right)\\,\\,\\,\\,\\,\\ne 0 \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & m+1\\ne 0 \\\\ & m-2\\ne 0 \\\\\\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne -1 \\\\ & m\\ne 2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span>","column":2}]}],"id_ques":38},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\sqrt{5}$","B. $2\\sqrt{5}$","C. $3\\sqrt{5}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/4.jpg' \/><\/center>Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$ cho \u0111i\u1ec3m $A (4; 2)$. \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n $OA$ l\u00e0 ?","hint":"H\u1ea1 $AH \\bot Ox$ t\u1ea1i $H$. <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago v\u00e0o tam gi\u00e1c vu\u00f4ng $OAH$. ","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv2/img\/D921_TB6.png' \/><\/center><br\/>H\u1ea1 $AH \\bot Ox$ t\u1ea1i $H$ $\\Rightarrow H (4; 0)$<br\/>Theo h\u00ecnh v\u1ebd ta c\u00f3 $AH= 2; OH =4$<br\/>X\u00e9t $\\Delta AHO$ vu\u00f4ng t\u1ea1i $H$ c\u00f3:<br\/> $\\begin{align} AO^2&=AH^2+OH^2\\,\\,(\u0110\u1ecbnh\\,\\,l\u00fd\\,\\,Py - ta - go)\\\\&=4+16\\\\&=20 \\\\ \\Rightarrow AO&= 2\\sqrt{5}\\\\ \\end{align}$ <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2\\sqrt{5}$.<\/span>"}]}],"id_ques":39},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $\\dfrac{6}{11}$","B. $\\dfrac{7}{11}$","C. $\\dfrac{8}{11}$"],"ques":"<span class='basic_left'> Cho h\u00e0m s\u1ed1 $f(x)= 5x - 3$ v\u00e0 $g(x) =-\\dfrac{1}{2}x +1$. T\u00ecm $a$ sao cho $f(a) =g(a)$. Gi\u00e1 tr\u1ecb c\u1ee7a $a$ t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 ?","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $f(a) = g(a)$ ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & f\\left( a \\right)=5a-3 \\\\ & g\\left( a \\right)=-\\dfrac{1}{2}a+1 \\\\ & f\\left( a \\right)=g\\left( a \\right)\\Leftrightarrow 5a-3=-\\dfrac{1}{2}a+1 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow \\dfrac{11}{2}a=4 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow a=\\dfrac{8}{11} \\\\ \\end{align}$<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $\\dfrac{8}{11}$. <\/span>"}]}],"id_ques":40}],"lesson":{"save":0,"level":2}}