{"segment":[{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang ","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u ","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Trong c\u00f9ng m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$, cho b\u1ed1n \u0111i\u1ec3m $A (-3; 4); B (-2; 1); C (1; 2)$ v\u00e0 $D (0; 5)$.<br\/><b> C\u00e2u 1: <\/b> T\u1ee9 gi\u00e1c $ABCD$ l\u00e0 h\u00ecnh g\u00ec?","select":["A. H\u00ecnh b\u00ecnh h\u00e0nh ","B. H\u00ecnh ch\u1eef nh\u1eadt","C. H\u00ecnh thoi ","D. H\u00ecnh vu\u00f4ng"],"hint":"T\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a t\u1ee9 gi\u00e1c r\u1ed3i d\u1ef1a v\u00e0o d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft t\u1ee9 gi\u00e1c \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh $ABCD$ l\u00e0 h\u00ecnh g\u00ec ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c t\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng \u0111\u1ec3 t\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng ch\u00e9o.<br\/> $BA=\\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}$<br\/>B\u01b0\u1edbc 2: D\u1ef1a theo d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft h\u00ecnh vu\u00f4ng \u0111\u1ec3 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/>$BA=\\sqrt{{{\\left( {{x}_{A}}-{{x}_{B}} \\right)}^{2}}+{{\\left( {{y}_{A}}-{{y}_{B}} \\right)}^{2}}}\\,$$=\\sqrt{{{\\left( -3+2 \\right)}^{2}}+{{\\left( 4-1 \\right)}^{2}}}\\,$$=\\sqrt{10}$<br\/>$CB=\\sqrt{{{\\left( {{x}_{B}}-{{x}_{C}} \\right)}^{2}}+{{\\left( {{y}_{B}}-{{y}_{C}} \\right)}^{2}}}\\,$$=\\sqrt{{{\\left( -2-1 \\right)}^{2}}+{{\\left( 1-2 \\right)}^{2}}}\\,$$=\\sqrt{10}$ <br\/>$DC=\\sqrt{{{\\left( {{x}_{C}}-{{x}_{D}} \\right)}^{2}}+{{\\left( {{y}_{C}}-{{y}_{D}} \\right)}^{2}}}\\,$$=\\sqrt{{{\\left( 1-0 \\right)}^{2}}+{{\\left( 2-5 \\right)}^{2}}}\\,$$=\\sqrt{10}$ <br\/>$DA=\\sqrt{{{\\left( {{x}_{A}}-{{x}_{D}} \\right)}^{2}}+{{\\left( {{y}_{A}}-{{y}_{D}} \\right)}^{2}}}\\,$$=\\sqrt{{{\\left( -3-0 \\right)}^{2}}+{{\\left( 4-5 \\right)}^{2}}}\\,$$=\\sqrt{10}$ <br\/>$CA=\\sqrt{{{\\left( {{x}_{A}}-{{x}_{C}} \\right)}^{2}}+{{\\left( {{y}_{A}}-{{y}_{C}} \\right)}^{2}}}\\,$$=\\sqrt{{{\\left( -3-1 \\right)}^{2}}+{{\\left( 4-2 \\right)}^{2}}}\\,$$=\\sqrt{20}=2\\sqrt{5}$<br\/>$DB=\\sqrt{{{\\left( {{x}_{B}}-{{x}_{D}} \\right)}^{2}}+{{\\left( {{y}_{B}}-{{y}_{D}} \\right)}^{2}}}\\,$$=\\sqrt{{{\\left( -2-0 \\right)}^{2}}+{{\\left( 1-5 \\right)}^{2}}}\\,$$=\\sqrt{20}=2\\sqrt{5}$<br\/>$\\Rightarrow AB=BC=CD=DA;\\,$$ AC=BD$<br\/>$\\Rightarrow$ T\u1ee9 gi\u00e1c $ABCD$ l\u00e0 h\u00ecnh vu\u00f4ng. <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":2}]}],"id_ques":191},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u ","temp":"fill_the_blank","correct":[[["-1"],["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Trong c\u00f9ng m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$, cho b\u1ed1n \u0111i\u1ec3m $A (-3; 4); B (-2; 1); C (1; 2)$ v\u00e0 $D (0; 5)$.<br\/><b> C\u00e2u 2: <\/b> $AC \\cap BD = I$. T\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m $I$ l\u00e0 (_input_; _input_)","hint":"$I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Trung \u0111i\u1ec3m $I$ c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $AB$ c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0 $I\\left( \\dfrac{x_A+x_B}{2}; \\dfrac{y_A+y_B}{2} \\right)$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$AC \\cap BD = I \\,\\Rightarrow I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$.<br\/>T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $I$ l\u00e0: <br\/>$\\left\\{ \\begin{aligned} & {{x}_{I}}=\\dfrac{{{x}_{A}}+{{x}_{C}}}{2} \\\\ & {{y}_{I}}=\\dfrac{{{y}_{A}}+{{y}_{C}}}{2} \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{I}}=\\dfrac{-3+1}{2}=-1 \\\\ & {{y}_{I}}=\\dfrac{4+2}{2}=3 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow I\\left( -1;3 \\right)$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-1$ v\u00e0 $3$. <\/span><\/span>"}]}],"id_ques":192},{"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","3"],["3","-2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv3/img\/9.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh : $|x+1|+|x-2|=5$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $x\\in $ {_input_; _input_}","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p v\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=|x+1|+|x-2|$. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh ngh\u0129a gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i $|x|=\\left\\{ \\begin{align} & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{x}\\ge \\text{0} \\\\ & -x\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{x}< \\text{0} \\\\\\end{align} \\right.$<br\/>Chia c\u00e1c tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed1i v\u1edbi $x$ \u0111\u1ec3 bi\u1ebft d\u1ea5u bi\u1ec3u th\u1ee9c trong tr\u1ecb tuy\u1ec7t \u0111\u1ed1i v\u00e0 b\u1ecf gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i d\u1ef1a v\u00e0o \u0111\u1ecbnh ngh\u0129a. Khi \u0111\u00f3 ta \u0111\u01b0\u1ee3c h\u00e0m s\u1ed1 vi\u1ebft d\u01b0\u1edbi d\u1ea1ng nhi\u1ec1u bi\u1ec3u th\u1ee9c. <br\/>V\u1ebd \u0111\u1ed3 th\u1ecb c\u00e1c h\u00e0m s\u1ed1 th\u00e0nh ph\u1ea7n \u1ee9ng v\u1edbi t\u1eebng tr\u01b0\u1eddng h\u1ee3p c\u1ee7a $x$ ta \u0111\u01b0\u1ee3c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00e3 cho. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\begin{aligned} & y=\\left| x+1 \\right|+\\left| x-2 \\right| \\\\ & \\left\\{ \\begin{aligned} & y=-2x+1\\,\\,\\,n\u1ebfu \\,\\,\\,x\\le -1 \\\\ & y=3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,n\u1ebfu\\,\\,\\,\\,-1 < x\\le 2 \\\\ & y=2x-1\\,\\,\\,\\,n\u1ebfu\\,\\,\\,x > 2 \\\\ \\end{aligned} \\right. \\\\\\end{aligned}$ <br\/>V\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng $y = -2x+1, y =2x-1$ v\u00e0 $y=3$ tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 . <br\/> \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 l\u00e0 \u0111\u01b0\u1eddng g\u1ea5p kh\u00fac li\u1ec1n n\u00e9t \u1edf h\u00ecnh d\u01b0\u1edbi<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv3/img\/D924_K2.png' \/><\/center><br\/>\u0110\u01b0\u1eddng th\u1eb3ng $y= 5$ c\u1eaft \u0111\u1ed3 th\u1ecb c\u1ee7a $y=|x+1|+|x-2|$ t\u1ea1i $M (-2; 5)$ v\u00e0 $N(3; 5)$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m $x_{1}=-2; x_{2}=3$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-2$ v\u00e0 $3$. <\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/> C\u00e1ch 2: Ta c\u00f3 th\u1ec3 gi\u1ea3i tr\u1ef1c ti\u1ebfp sau khi chia kho\u1ea3ng m\u00e0 kh\u00f4ng c\u1ea7n v\u1ebd \u0111\u1ed3 th\u1ecb<\/span>"}]}],"id_ques":193},{"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}{2}$","B. $\\dfrac{1}{3}$","C. $\\dfrac{1}{4}$"],"ques":"T\u00ecm $m$ \u0111\u1ec3 \u0111\u1ed3 th\u1ecb hai h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t sau: <br\/>$mx+(m-1)y-2(m+2)=0$$\\,\\,\\,\\,\\,\\,\\,\\,\\,(1)$<br\/>$3mx-(3m+1)y-(5m+4)=0$$\\,\\,\\,\\,\\,(2)$<br\/>song song v\u1edbi nhau <br\/>","hint":": Bi\u1ebfn \u0111\u1ed5i hai h\u00e0m s\u1ed1 v\u1ec1 d\u1ea1ng $y= ax + b$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i hai h\u00e0m s\u1ed1 v\u1ec1 d\u1ea1ng $y=ax + b$.<br\/>B\u01b0\u1edbc 2: $(d)$ song song v\u1edbi $(d')$ khi h\u1ec7 s\u1ed1 g\u00f3c b\u1eb1ng nhau v\u00e0 tung \u0111\u1ed9 g\u1ed1c kh\u00e1c nhau. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$y=\\dfrac{-m}{m-1}x+\\dfrac{2\\left( m+2 \\right)}{m-1}$$\\,\\,\\,\\,\\,\\,\\,\\left( m\\ne 1 \\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right)$<br\/>$y=\\dfrac{3m}{3m+1}x-\\dfrac{\\left( 5m+4 \\right)}{3m+1}$$\\,\\,\\,\\,\\left( m\\ne -\\dfrac{1}{3} \\right)\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right)$<br\/>Hai \u0111\u01b0\u1eddng th\u1eb3ng n\u00e0y song song v\u1edbi nhau khi <br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\left\\{ \\begin{aligned} & \\dfrac{-m}{m-1}=\\dfrac{3m}{3m+1} \\\\ & \\dfrac{2\\left( m+2 \\right)}{\\left( m-1 \\right)}\\ne -\\dfrac{5m+4}{3m+1} \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & m\\left( \\dfrac{3}{3m+1}+\\dfrac{1}{m-1} \\right)=0 \\\\ & \\left( 4m+4 \\right)\\left( 3m+1 \\right)\\ne \\left( 5m+4 \\right)\\left( -m+1 \\right) \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & m\\left( 6m-2 \\right)=0 \\\\ & 17{{m}^{2}}+17m\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & \\left[ \\begin{aligned} & m=0 \\\\ & m=\\dfrac{1}{3} \\\\ \\end{aligned} \\right. \\\\ & \\left[ \\begin{aligned} & m\\ne 0 \\\\ & m\\ne -1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.\\Leftrightarrow m=\\dfrac{1}{3} \\\\ \\end{aligned}$ <br\/>"}]}],"id_ques":194},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-6"],["4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv3/img\/4.jpg' \/><\/center>Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$ cho c\u00e1c \u0111i\u1ec3m $A(-6; 4), B (-2; -2)$ v\u00e0 $C (-3; 6)$. <br\/>T\u1ecda \u0111\u1ed9 tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$ c\u00f3 ho\u00e0nh \u0111\u1ed9 v\u00e0 tung \u0111\u1ed9 l\u1ea7n l\u01b0\u1ee3t l\u00e0 _input_ v\u00e0 _input_","hint":"T\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c \u0111o\u1ea1n th\u1eb3ng $AB, BC, AC$ r\u1ed3i nh\u1eadn d\u1ea1ng tam gi\u00e1c \u0111\u00e3 cho.<br\/>\u00c1p d\u1ee5ng $AB=\\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}$","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$AB=\\sqrt{{{\\left( {{x}_{A}}-{{x}_{B}} \\right)}^{2}}+{{\\left( {{y}_{A}}-{{y}_{B}} \\right)}^{2}}}\\,$$=\\sqrt{{{\\left( -6+2 \\right)}^{2}}+{{\\left( 4+2 \\right)}^{2}}}\\,$$=\\sqrt{52}=2\\sqrt{13}$ <br\/>$AC=\\sqrt{{{\\left( {{x}_{A}}-{{x}_{C}} \\right)}^{2}}+{{\\left( {{y}_{A}}-{{y}_{C}} \\right)}^{2}}}\\,$$=\\sqrt{{{\\left( -6+3 \\right)}^{2}}+{{\\left( 4-6 \\right)}^{2}}}\\,$$=\\sqrt{13}$<br\/>$BC=\\sqrt{{{\\left( {{x}_{B}}-{{x}_{C}} \\right)}^{2}}+{{\\left( {{y}_{B}}-{{y}_{C}} \\right)}^{2}}}\\,$$=\\sqrt{{{\\left( -2+3 \\right)}^{2}}+{{\\left( -2-6 \\right)}^{2}}}\\,$$=\\sqrt{65}$<br\/>$\\Rightarrow B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\\Rightarrow \\Delta ABC$ vu\u00f4ng t\u1ea1i $A$ (\u0111\u1ecbnh l\u00ed \u0111\u1ea3o Pytago) <br\/>Do \u0111\u00f3 t\u1ecda \u0111\u1ed9 tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$ ch\u00ednh l\u00e0 \u0111i\u1ec3m $A$. <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-6$ v\u00e0 $4$.<\/span> <\/span> "}]}],"id_ques":195},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Trong c\u00f9ng m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$, cho ba \u0111i\u1ec3m $A (2; 3), B (-1; -3)$ v\u00e0 $C \\left(\\dfrac{1}{2}; 0\\right)$<br\/>Ba \u0111i\u1ec3m $A, B, C$ th\u1eb3ng h\u00e0ng.","select":["A. \u0110\u00fang","B. Sai"],"hint":"\u0110i\u1ec3m $C$ c\u00f3 thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $AB$ kh\u00f4ng? ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$. <br\/><b> B\u01b0\u1edbc 2:<\/b> Ki\u1ec3m tra \u0111i\u1ec3m $C$ c\u00f3 thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $AB$ kh\u00f4ng.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align} & AB:\\,\\,\\,y=ax+b \\\\ & A\\left( 2;3 \\right)\\in AB \\Rightarrow 2a+b=3\\,\\,\\left( 1 \\right) \\\\ \\end{align}$ <br\/>$B\\left( -1;-3 \\right)\\Rightarrow -a+b=-3\\,$$\\Leftrightarrow b=-3+a\\,\\,\\,\\left( 2 \\right)$<br\/>Thay (2) v\u00e0o (1) ta \u0111\u01b0\u1ee3c : $2a-3+a=3\\Leftrightarrow a=2\\Rightarrow b=-1$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ l\u00e0: $y= 2x \u2013 1$<br\/>Ta c\u00f3: $x=\\dfrac{1}{2}$ th\u00ec $y=2.\\dfrac{1}{2}-1=0$ <br\/>Do \u0111\u00f3 $\\,C\\in AB\\Rightarrow A, B, C$ th\u1eb3ng h\u00e0ng. <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 A.<\/span><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span> B\u00e0i to\u00e1n n\u00e0y c\u00f3 th\u1ec3 ch\u1ee9ng minh b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p kh\u00e1c<br\/><b> C\u00e1ch 2: <\/b>Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ v\u00e0 $BC$. Ki\u1ec3m tra xem hai \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 tr\u00f9ng nhau kh\u00f4ng?<br\/><b> C\u00e1ch 3:<\/b> T\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh $AB, AC$ v\u00e0 $BC$.<br\/> $AB=AC+BC \\,\\Rightarrow C$ n\u1eb1m gi\u1eefa $A$ v\u00e0 $B \\,\\Rightarrow A, B, C$ th\u1eb3ng h\u00e0ng.<\/span>","column":2}]}],"id_ques":196},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["0","-1"],["-1","0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\,y=(2m+1)x-2\\,\\,\\,\\left(m \\ne -\\dfrac{1}{2}\\right)$<br\/>$(d)\\cap Ox = A,\\,\\, (d) \\cap Oy = B$<br\/><span class='basic_left'><b> C\u00e2u 1: <\/b> T\u00ecm $m$ sao cho kho\u1ea3ng c\u00e1ch t\u1eeb g\u1ed1c t\u1ecda \u0111\u1ed9 $O$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ b\u1eb1ng $\\sqrt{2}$.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $\\,m \\in${_input_; _input_}<\/span>","hint":"D\u00f9ng h\u1ec7 th\u1ee9c gi\u1eefa c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao trong tam gi\u00e1c vu\u00f4ng <br\/>$\\dfrac{1}{OH^2}=\\dfrac{1}{OA^2}+\\dfrac{1}{OB^2}$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b> B\u01b0\u1edbc 1: <\/b> T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $A$ v\u00e0 $B$ theo $m$. <br\/><b> B\u01b0\u1edbc 2:<\/b> D\u00f9ng h\u1ec7 th\u1ee9c gi\u1eefa c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao trong tam gi\u00e1c vu\u00f4ng \u0111\u1ec3 t\u00ecm $m$. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{aligned} & \\left( d \\right)\\cap \\text{Ox=A}\\Rightarrow \\text{A}\\left( \\dfrac{2}{2m+1};0 \\right) \\\\ & \\left( d \\right)\\cap \\text{Oy=B}\\Rightarrow B\\left( 0;-2 \\right) \\\\ & \\Rightarrow OA=\\frac{2}{|2m+1|};\\,\\,OB=2 \\\\ \\end{aligned}$ <br\/>Kho\u1ea3ng c\u00e1ch t\u1eeb g\u1ed1c t\u1ecda \u0111\u1ed9 $O$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $d$ b\u1eb1ng $\\sqrt{2}$<br\/>$\\begin{aligned} & \\Rightarrow \\dfrac{1}{O{{A}^{2}}}+\\dfrac{1}{O{{B}^{2}}}={{\\left( \\frac{1}{\\sqrt{2}} \\right)}^{2}} \\\\ & \\Leftrightarrow \\dfrac{{{\\left( 2m+1 \\right)}^{2}}}{4}+\\dfrac{1}{4}=\\dfrac{1}{2} \\\\ & \\Leftrightarrow {{\\left( 2m+1 \\right)}^{2}}=1 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2m+1=1 \\\\ & 2m+1=-1 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & m=0 \\\\ & m=-1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $0$ v\u00e0 $-1$<\/span>"}]}],"id_ques":197},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\left[ \\begin{aligned} & m=\\dfrac{3}{2} \\\\ & m=-\\dfrac{5}{2} \\\\ \\end{aligned} \\right.$","B. $\\left[ \\begin{aligned} & m=\\dfrac{1}{2} \\\\ & m=-\\dfrac{5}{2} \\\\ \\end{aligned} \\right.$","C. $\\left[ \\begin{aligned} & m=\\dfrac{1}{2} \\\\ & m=-\\dfrac{3}{2} \\\\ \\end{aligned} \\right.$"],"ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\,y=(2m+1)x-2\\,\\,\\,\\left(m \\ne -\\dfrac{1}{2}\\right)$<br\/>$(d)\\cap Ox = A,\\,\\, (d) \\cap Oy = B$<br\/><span class='basic_left'>T\u00ecm $m$ sao cho di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ b\u1eb1ng $\\dfrac{1}{2}$<\/span>","hint":"T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $AOB$ theo $m$. ","explain":"<span class='basic_left'> Theo c\u00e2u tr\u00ean, ta c\u00f3: $\\Rightarrow OA=\\dfrac{2}{|2m+1|};\\,\\,OB=2 $ <br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ b\u1eb1ng $\\dfrac{1}{2}$<br\/>$\\begin{aligned} & \\Rightarrow \\dfrac{OA.OB}{2}\\,\\,\\,\\,\\,\\,=\\dfrac{1}{2} \\\\ & \\Leftrightarrow \\dfrac{2}{\\left| 2m+1 \\right|}.2=1 \\\\ & \\Leftrightarrow \\left| 2m+1 \\right|\\,\\,\\,\\,\\,\\,=4 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2m+1=4 \\\\ & 2m+1=-4 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & m=\\dfrac{3}{2} \\\\ & m=-\\dfrac{5}{2} \\\\ \\end{aligned} \\right. \\\\\\end{aligned}$ <br\/>"}]}],"id_ques":198},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $y_{min} = \\dfrac{1}{2}$ khi $x = -1$","B. $y_{min} = \\dfrac{1}{2}$ khi $x = 0$","C. $y_{min} = \\dfrac{1}{2}$ khi $x = 1$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv3/img\/2.png' \/><\/center>T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 $y=\\dfrac{{{x}^{2}}+2x+2}{{{x}^{2}}+2x+3}$ <br\/>\u0110\u00e1p s\u1ed1: $y_{min}$= ? khi $x=$ ?","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c th\u00e0nh m\u1ed9t s\u1ed1 th\u1ef1c c\u1ed9ng v\u1edbi m\u1ed9t ph\u00e2n th\u1ee9c c\u00f3 t\u1eed l\u00e0 s\u1ed1 nguy\u00ean. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c th\u00e0nh m\u1ed9t s\u1ed1 th\u1ef1c c\u1ed9ng v\u1edbi m\u1ed9t ph\u00e2n th\u1ee9c c\u00f3 t\u1eed l\u00e0 s\u1ed1 nguy\u00ean<br\/>B\u01b0\u1edbc 2: \u0110\u00e1nh gi\u00e1 bi\u1ec3u th\u1ee9c v\u1eeba bi\u1ebfn \u0111\u1ed5i. \u00c1p d\u1ee5ng: $\\dfrac{a}{[f(x)]^2+b}\\le \\dfrac{a}{b}$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\begin{align}y & =\\dfrac{{{x}^{2}}+2x+2}{{{x}^{2}}+2x+3} \\\\ & =\\dfrac{\\left( {{x}^{2}}+2x+3 \\right)-1}{{{x}^{2}}+2x+3} \\\\ & =1-\\dfrac{1}{{{x}^{2}}+2x+3} \\\\ & =1-\\dfrac{1}{{{\\left( x+1 \\right)}^{2}}+2} \\\\ \\end{align}$<br\/>Ta c\u00f3 $\\dfrac{1}{{{\\left( x+1 \\right)}^{2}}+2}\\le \\dfrac{1}{2}$<br\/>$\\Rightarrow y= 1-\\dfrac{1}{{{\\left( x+1 \\right)}^{2}}+2}\\,$$\\ge 1-\\dfrac{1}{2}=\\dfrac{1}{2}$ <br\/>$\\Rightarrow {{y}_{\\min }}=\\dfrac{1}{2}\\Leftrightarrow {{\\left( x+1 \\right)}^{2}}=0\\,$$\\Leftrightarrow x=-1$<br\/> <span class='basic_pink'><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span> <br\/>\u0110\u1ed1i v\u1edbi b\u00e0i t\u00ecm GTLN v\u00e0 GTNN c\u1ee7a ph\u00e2n th\u1ee9c c\u00f3 b\u1eadc c\u1ee7a t\u1eed v\u00e0 b\u1eadc c\u1ee7a m\u1eabu b\u1eb1ng nhau, ta c\u00f3 th\u1ec3 th\u1ef1c hi\u1ec7n chia t\u1eed th\u1ee9c cho m\u1eabu th\u1ee9c \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c \u0111\u00f3 th\u00e0nh m\u1ed9t s\u1ed1 th\u1ef1c c\u1ed9ng v\u1edbi m\u1ed9t ph\u00e2n th\u1ee9c c\u00f3 t\u1eed l\u00e0 s\u1ed1 nguy\u00ean.<\/span> "}]}],"id_ques":199},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 $y=\\sqrt{2{{x}^{2}}-3x-2}\\,$$+\\dfrac{3-2x}{\\sqrt{x-1}}$ l\u00e0:","select":["A. $x >2$","B. $x > 1$","C. $x \\ge 1$ ","D. $ x \\ge 2$"],"hint":"<br\/>+ $\\sqrt{A}$ c\u00f3 ngh\u0129a $\\Leftrightarrow A \\ge 0$<br\/>+ $\\dfrac{a}{\\sqrt{b}}$ c\u00f3 ngh\u0129a $\\Leftrightarrow b > 0$ ","explain":"<span class='basic_left'>Ta c\u00f3 $y=\\sqrt{2{{x}^{2}}-3x-2}\\,$$+\\dfrac{3-2x}{\\sqrt{x-1}}\\,$$=\\sqrt{\\left( x-2 \\right)\\left( 2x+1 \\right)}+\\dfrac{3-2x}{\\sqrt{x-1}}$<br\/>\u0110i\u1ec1u ki\u1ec7n: <br\/>$\\left\\{ \\begin{aligned} & \\left( x-2 \\right)\\left( 2x+1 \\right)\\ge 0 \\\\ & x-1>0 \\\\ \\end{aligned} \\right.\\,$ $\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left[ \\begin{aligned} & \\left\\{ \\begin{aligned} & x-2\\ge 0 \\\\ & 2x+1\\ge 0 \\\\ \\end{aligned} \\right. \\\\ & \\left\\{ \\begin{aligned} & x-2\\le 0 \\\\ & 2x+1\\le 0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right. \\\\ & x>1 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left[ \\begin{aligned} & x\\ge 2 \\\\ & x\\le -\\dfrac{1}{2} \\\\ \\end{aligned} \\right. \\\\ & x>1 \\\\ \\end{aligned} \\right.\\Leftrightarrow x\\ge 2$<br\/> Do \u0111\u00f3 \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 $x\\ge 2$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":200}],"lesson":{"save":0,"level":3}}