{"segment":[{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$, cho $A(1; 1)$ v\u00e0 $B(3; 9)$. Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $AB$ l\u00e0: ","select":["A. $ y=\\dfrac{1}{4}x+\\dfrac{11}{2}$ ","B. $y=-\\dfrac{1}{4}x+\\dfrac{11}{2}$ ","C. $y=-\\dfrac{1}{4}x-\\dfrac{11}{2}$ ","D. $y=\\dfrac{1}{4}x-\\dfrac{11}{2}$ "],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> T\u00ecm ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$.<br\/><b> B\u01b0\u1edbc 2: <\/b> T\u00ecm trung \u0111i\u1ec3m \u0111o\u1ea1n th\u1eb3ng $AB$.<br\/><b> B\u01b0\u1edbc 3:<\/b> Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi $AB$ t\u1ea1i trung \u0111i\u1ec3m \u0111\u00f3.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ c\u00f3 d\u1ea1ng: $y=ax + b$ <br\/>$A\\left( 1;1 \\right)\\in AB\\Rightarrow a.1+b=1$$\\,\\Leftrightarrow b=1-a\\,\\,\\,\\,\\,\\left( 1 \\right)$ <br\/>$B\\left( 3;9 \\right)\\in AB$$\\,\\Rightarrow 3.a+b=9\\,\\,\\,\\left( 2 \\right)$<br\/>Thay (1) v\u00e0o (2) ta \u0111\u01b0\u1ee3c: $3a+1-a=9\\Leftrightarrow a=4\\Rightarrow b=-3$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$: $y= 4x \u2013 3 \\Rightarrow $ c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c $a= 4$.<br\/>G\u1ecdi $(d):\\, y=a'x+b'$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$<br\/>$ \\Rightarrow d\\bot AB\\Rightarrow a.a'=-1\\Rightarrow a'=\\dfrac {-1}{4}\\Rightarrow \\left( d \\right):\\,\\,\\,y=-\\dfrac{1}{4}x+b'$ <br\/>G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $AB$. <br\/>$\\left\\{ \\begin{align} & {{x}_{H}}=\\dfrac{{{x}_{A}}+{{x}_{B}}}{2} \\\\ & {{y}_{H}}=\\dfrac{{{y}_{A}}+{{y}_{B}}}{2} \\\\ \\end{align} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{align} & {{x}_{H}}=\\dfrac{1+3}{2}=2 \\\\ & {{y}_{H}}=\\dfrac{1+9}{2}=5 \\\\ \\end{align} \\right.\\,$$\\Leftrightarrow H\\left( 2;5 \\right)$ <br\/>M\u00e0 $H\\in \\left( d \\right)\\,$$\\Rightarrow 5=-\\dfrac{1}{4}.2+b'\\Leftrightarrow b'=\\dfrac{11}{2}$<br\/>$\\Rightarrow \\left( d \\right):\\,\\,\\,y=-\\dfrac{1}{4}x+\\dfrac{11}{2}$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 B.<\/span><br\/><b> L\u01b0u \u00fd: <\/b><br\/>+ Hai \u0111\u01b0\u1eddng th\u1eb3ng $y=ax+b$ v\u1edbi ($a\\ne 0$), $y=a'x+b'$ v\u1edbi ($a'\\ne 0$) vu\u00f4ng g\u00f3c v\u1edbi nhau khi $a.a'=-1$.<br\/>+ C\u00f4ng th\u1ee9c trung \u0111i\u1ec3m $M(x_M;y_M)$ c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $AB$ v\u1edbi $A(x_A;y_A)$ v\u00e0 $B(x_B;y_B)$ l\u00e0:<br\/>$\\left\\{ \\begin{align} & {{x}_{M}}=\\dfrac{{{x}_{A}}+{{x}_{B}}}{2} \\\\ & {{y}_{M}}=\\dfrac{{{y}_{A}}+{{y}_{B}}}{2} \\\\ \\end{align} \\right.$<\/span>","column":2}]}],"id_ques":141},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho hai \u0111\u01b0\u1eddng th\u1eb3ng $\\,(d_{1}):\\,\\,y=3x+5m+2\\,\\,$ v\u00e0 $\\,(d_{2}):\\,\\,y=7x-3m-6$<br\/>T\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m $A$ c\u1ee7a $\\,(d_{1})\\,$ v\u00e0 $\\,(d_{2})$ l\u00e0:","select":["A. $ A(m-1; 8m - 1)$","B. $A(m+1; 8m + 5)$","C. $ A(2m-2; 11m - 4)$","D. $ A(2m+2; 11m + 8)$"],"explain":"<span class='basic_left'> $\\,(d_{1}):\\,\\,y=3x+5m+2\\,\\,$ c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c $a= 3$<br\/>$\\,(d_{2}):\\,\\,y=7x-3m-6$ c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c $a'=7$<br\/>$\\Rightarrow a \\ne a' \\Rightarrow d_{1}\\cap d_{2}=A (x_{1};y_{1})$<br\/>Ho\u00e0nh \u0111\u1ed9 \u0111i\u1ec3m $A$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned}\\,\\,\\,\\,\\,3x_{1}+5m+2&=7x_{1}-3m-6\\\\ \\Leftrightarrow 4x_{1}&=8m+8\\\\ \\Leftrightarrow \\,\\,\\,x_{1}&=2m+2\\\\ \\end{aligned}$<br\/>$\\Rightarrow y_{1}=11m +8\\,$$ \\Rightarrow A(2m+2; 11m + 8)$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":142},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ song song v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $x+2y= 1$ v\u00e0 \u0111i qua giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng $\\,(d_{1}):\\,\\,2x-3y=4\\,$ v\u00e0 $\\,(d_{2}):\\,\\,3x+y=5$ c\u00f3 d\u1ea1ng: ","select":["A. $y=\\dfrac{1}{2}x+\\dfrac{15}{22}$","B. $y=-\\dfrac{1}{2}x+\\dfrac{15}{22}$","C. $y=-\\dfrac{1}{2}x-\\dfrac{15}{22}$","D. $y=\\dfrac{1}{2}x-\\dfrac{15}{22}$"],"hint":"T\u00ecm giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng $(d_{1})$ v\u00e0 $(d_{2})$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<br\/><\/span><b>B\u01b0\u1edbc 1: <\/b> T\u00ednh h\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\, y=ax+ b$<br\/><b> B\u01b0\u1edbc 2: <\/b> T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $(d_{1})$ v\u00e0 $(d_{2})$<br\/><b> B\u01b0\u1edbc 3: <\/b> Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\, y= ax + b$ \u0111\u1ec3 t\u00ecm ra $b$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110\u01b0\u1eddng th\u1eb3ng $x+2y=1\\Leftrightarrow y=-\\dfrac{1}{2}x+\\dfrac{1}{2}$ n\u00ean c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c l\u00e0 $-\\dfrac{1}{2}$<br\/>\u0110\u01b0\u1eddng th\u1eb3ng $(d):\\,y=ax+b$ song song v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $x+2y=1\\,$$\\Rightarrow \\left( d \\right):\\,\\,y=-\\dfrac{1}{2}x+b\\,\\,\\,\\left( b\\ne \\dfrac{1}{2} \\right)$ <br\/>$\\left( {{d}_{1}} \\right):\\,\\,2x-3y=4\\,$$\\Leftrightarrow \\left( {{d}_{1}} \\right):\\,\\,\\,y=\\dfrac{2}{3}x-\\dfrac{4}{3}$ <br\/>$\\left( {{d}_{2}} \\right):3x+y=5\\,$$\\Leftrightarrow \\,\\left( {{d}_{2}} \\right):\\,\\,\\,y=-3x+5$ <br\/>$\\left( {{d}_{1}} \\right)\\cap \\left( {{d}_{2}} \\right)=M.\\,$ <br\/> Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m $M$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\dfrac{2}{3}x-\\dfrac{4}{3}=-3x+5\\,$<br\/>$\\Leftrightarrow x=\\dfrac{19}{11}$<br\/>$\\Rightarrow y=-\\dfrac{2}{11}$ <br\/>$\\Rightarrow M(\\dfrac{19}{11};-\\dfrac{2}{11})$<br\/>V\u00ec $M\\in \\left( d \\right)\\Rightarrow -\\dfrac{2}{11}=-\\dfrac{19}{22}+b\\,$$\\Leftrightarrow b=\\dfrac{15}{22}\\ne \\dfrac{1}{2}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ c\u00f3 d\u1ea1ng: $y=-\\dfrac{1}{2}x+\\dfrac{15}{22}$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":143},{"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","-2"],["-2","-1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\,y=(m-2)x+ 3$ $\\,(m\\ne 2)$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $(d'):\\,y=-m^2x+1$ $\\,(m\\ne 0)$ <\/span><br\/><span class='basic_left'><b> C\u00e2u 1:<\/b> T\u00ecm $m$ \u0111\u1ec3 $d\/\/d'$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $m\\in\\,$ {_input_; _input_}<\/span>","explain":" $\\begin{aligned} & d\/\/d'\\Leftrightarrow m-2=-{{m}^{2}} \\\\ & \\Leftrightarrow {{m}^{2}}+m-2\\,\\,\\,\\,\\,\\,\\,\\,\\,=0 \\\\ & \\Leftrightarrow {{m}^{2}}+2m-m-2\\,\\,\\,\\,\\,\\,\\,\\,\\,=0 \\\\ & \\Leftrightarrow m(m+2)-(m+2)\\,\\,\\,\\,\\,\\,\\,\\,\\,=0 \\\\ & \\Leftrightarrow \\left( m-1 \\right)\\left( m+2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & m-1=0 \\\\ & m+2=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & m=1 \\\\ & m=-2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$ v\u00e0 $-2$.<\/span>"}]}],"id_ques":144},{"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":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\,y=(m-2)x+ 3$ $\\,(m\\ne 2)$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $(d'):\\,y=-m^2x+1$ $\\,(m\\ne 0)$ <\/span><br\/><span class='basic_left'><b> C\u00e2u 2:<\/b> T\u00ecm $m$ \u0111\u1ec3 $(d)$ c\u1eaft $Ox$ t\u1ea1i $A$, c\u1eaft $Oy$ t\u1ea1i $B$ m\u00e0 $\\widehat{BAO}={{60}^{0}}$.<\/span> ","select":["A. $m=2-\\sqrt{3}$","B. $m=4- \\sqrt{3}$","C. $m=2\\pm \\sqrt{3}$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<br\/><\/span><b>B\u01b0\u1edbc 1: <\/b> T\u00ednh t\u1ecda \u0111\u1ed9 $A, B$ theo $m$.<br\/><b> B\u01b0\u1edbc 2: <\/b> D\u1ef1a v\u00e0o \u0111\u1ed3 th\u1ecb \u0111\u1ec3 t\u00ednh $tg\\,\\widehat{BAO}$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai11/lv3/img\/D923_K1.png' \/><\/center>$d\\cap Ox=A\\Rightarrow A\\left( \\dfrac{-3}{m-2};0 \\right);\\,\\,$$\\,\\,\\,\\,OA=\\dfrac{3}{|m-2|} $ <br\/> $ d\\cap Oy=B$ n\u00ean $B$ c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng $0$. <br\/> Do \u0111\u00f3 ta thay $x=0$ v\u00e0o $(d)$: $y=3\\Rightarrow B\\left( 0;\\,\\,3 \\right) $<br\/>X\u00e9t $\\,\\Delta AOB\\,$ c\u00f3: $ \\widehat{BAO}={{60}^{0}}\\Leftrightarrow tg \\,\\widehat{BAO}=\\sqrt{3} $<br\/> $ \\Leftrightarrow \\dfrac{OB}{OA}=\\sqrt{3}\\Leftrightarrow \\dfrac{3}{\\dfrac{3}{|m-2|}}=\\sqrt{3}$$\\,\\,\\Leftrightarrow \\left| m-2 \\right|=\\sqrt{3}\\Leftrightarrow m=2\\pm \\sqrt{3}$<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":3}]}],"id_ques":145},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$, cho $A (-5; -1); B(-1; -4); C(3; 2)$. Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng cao xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh $B$ c\u1ee7a tam gi\u00e1c $ABC$ c\u00f3 d\u1ea1ng:","select":["A. $y=-\\dfrac{8}{3}x-\\dfrac{20}{3}$","B. $y=\\dfrac{8}{3}x+\\dfrac{20}{3}$","C. $y=-\\dfrac{8}{3}x+\\dfrac{20}{3}$","D. $y=\\dfrac{8}{3}x-\\dfrac{20}{3}$ "],"hint":"\u0110\u01b0\u1eddng cao $BH$ \u0111i qua $B$ v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $AC$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<br\/><\/span><b>B\u01b0\u1edbc 1: <\/b> Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AC$.<br\/><b> B\u01b0\u1edbc 2: <\/b> K\u1ebb $BH \\bot AC$ t\u1ea1i $H$. Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng cao $BH$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AC$ c\u00f3 d\u1ea1ng: $y= ax+ b$<br\/>$A(-5; -1)\\in AC\\,\\Rightarrow -5a + b =-1 \\Leftrightarrow b=5a -1\\,\\,\\,\\,(1)$<br\/>$C(3; 2)\\in AC$$\\,\\Rightarrow 3a + b =2 \\,\\,\\,\\,\\,(2)$<br\/>Thay (1) v\u00e0o (2) ta \u0111\u01b0\u1ee3c: $\\,3a + 5a - 1 = 2\\Leftrightarrow a= \\dfrac{3}{8}\\Rightarrow b= \\dfrac{7}{8}$<br\/> Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AC:\\,\\,\\,y=\\dfrac{3}{8}x+\\dfrac{7}{8}$<br\/>K\u1ebb $BH \\bot AC$ t\u1ea1i $ H$.<br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $BH$ c\u00f3 d\u1ea1ng: $y= a'x + b'$ <br\/>V\u00ec $BH \\bot AC$ n\u00ean $a.a'=-1\\Rightarrow \\dfrac{3}{8}.a'=-1 \\Rightarrow a'= -\\dfrac{8}{3}$<br\/>$B(-1; -4) \\in BH \\Rightarrow a'.(-1)+b'=-4 $$\\,\\Leftrightarrow \\dfrac{8}{3}+b'=-4\\Leftrightarrow b'=-\\dfrac{20}{3}$<br\/>$\\Rightarrow BH : y=-\\dfrac{8}{3}x-\\dfrac{20}{3}$<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":146},{"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":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$ cho hai \u0111\u01b0\u1eddng th\u1eb3ng $y= 2- 2x$ v\u00e0 $y = x + 3a + 5$ (v\u1edbi $a$ l\u00e0 tham s\u1ed1) c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m $A(x_{o}; y_{o})$ th\u1ecfa m\u00e3n $x_{o}^{2}+y_{o}^{2}=40$. Khi \u0111\u00f3 gi\u00e1 tr\u1ecb nguy\u00ean d\u01b0\u01a1ng c\u1ee7a $a$ l\u00e0 _input_","hint":"T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $A$ theo $a$.","explain":" <span class='basic_left'> ${{d}_{1}}\\cap {{d}_{2}}=A.$ <br\/> Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m $A$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\begin{aligned} & \\,\\,\\,\\,2-2{{x}_{o}}={{x}_{o}}+3a+5 \\\\ & \\Leftrightarrow 3{{x}_{o}}=-3a-3 \\\\ & \\Leftrightarrow {{x}_{o}}=-a-1 \\\\ & \\Rightarrow {{y}_{o}}=2-2x_o=2+2a+2=2a+4 \\\\ \\end{aligned}$ <br\/>$\\Rightarrow A\\left( -a-1;2a+4 \\right)$ <br\/>Theo \u0111\u1ea7u b\u00e0i ta c\u00f3: <br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,x_{o}^{2}+y_{o}^{2}=40 \\\\ & \\Leftrightarrow {{\\left( -a-1 \\right)}^{2}}+{{\\left( 2a+4 \\right)}^{2}}=40 \\\\ & \\Leftrightarrow {{a}^{2}}+2a+1+4{{a}^{2}}+16a+16=40 \\\\ & \\Leftrightarrow 5{{a}^{2}}+18a-23=0 \\\\ & \\Leftrightarrow 5{{a}^{2}}+23a-5a-23=0\\\\ & \\Leftrightarrow a(5a+23)-(5a+23)=0 \\\\ & \\Leftrightarrow \\left( a-1 \\right)\\left( 5a+23 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & a=1 \\\\ & a=-\\dfrac{23}{5} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$.<\/span><\/span>"}]}],"id_ques":147},{"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"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho hai \u0111\u01b0\u1eddng th\u1eb3ng $(d_{1}):\\,y= x+ 2$; $\\,\\,\\,\\,$$(d_{2}):\\,y= 2x + 1$ $\\,\\,\\,\\,$$(d_{3}):\\,y=(m^2+1)x+m$<br\/><b> C\u00e2u 1: <\/b> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 th\u00ec $\\, d_{3}\/\/d_{2}$<br\/><b> \u0110\u00e1p s\u1ed1 : <\/b> $m=$ _input_<\/span>","hint":"Cho hai \u0111\u01b0\u1eddng th\u1eb3ng $y= ax + b\\, (d)$ v\u00e0 $y= a'x + b'\\, (d')$, ($\\,a\\ne 0 ;\\,a'\\ne 0\\,$)<br\/>(d) \/\/ (d') $\\,\\Leftrightarrow a= a'; b\\ne b'$","explain":" <span class='basic_left'>\u0110\u1ec3 $\\left( {{d}_{3}} \\right)\/\/\\left( {{d}_{2}} \\right)\\Leftrightarrow \\left\\{ \\begin{aligned} & {{m}^{2}}+1=2 \\\\ & m\\ne 1 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{m}^{2}}-1=0 \\\\ & m\\ne 1 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left[ \\begin{aligned} & m=1 \\\\ & m=-1 \\\\ \\end{aligned} \\right. \\\\ & m\\ne 1 \\\\ \\end{aligned} \\right.\\Leftrightarrow m=-1$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-1$<\/span><br\/><br\/><span class='basic_green'>L\u01b0u \u00fd: <\/span> H\u1ecdc sinh hay qu\u00ean \u0111i\u1ec1u ki\u1ec7n $\\,b\\ne b'$<\/span><\/span>"}]}],"id_ques":148},{"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":[[["-2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho hai \u0111\u01b0\u1eddng th\u1eb3ng $(d_{1}):\\,y= x+ 2$; $\\,\\,\\,\\,$$(d_{2}):\\,y= 2x + 1$ $\\,\\,\\,\\,$$(d_{3}):\\,y=(m^2+1)x+m$<br\/><b> C\u00e2u 2: <\/b> T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 ba \u0111\u01b0\u1eddng th\u1eb3ng tr\u00ean c\u1eaft nhau t\u1ea1i m\u1ed9t \u0111i\u1ec3m. <br\/><b> \u0110\u00e1p s\u1ed1 : <\/b> $m=$ _input_<\/span>","hint":"T\u00ecm giao \u0111i\u1ec3m c\u1ee7a $(d_{1})$ v\u00e0 $(d_{2})$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng $(d_{1})$ v\u00e0 $(d_{2})$<br\/>B\u01b0\u1edbc 2: Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o $\\,(d_{3})$ \u0111\u1ec3 t\u00ednh ra $m$.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$d_{1} \\cap d_{2}=A.$ <br\/> Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m $A$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$x+2= 2x +1 \\Leftrightarrow x= 1 \\Rightarrow y=3 $$\\,\\,\\,\\,\\,$$\\Rightarrow A(1;3)$ <br\/>Ba \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m $A(1; 3)$<br\/>$\\left\\{ \\begin{aligned} & d_1 \\text {c\u1eaft}\\,d_3 \\\\ & d_2 \\text {c\u1eaft}\\,d_3 \\\\ & A(1;3)\\in d_3 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow\\left\\{ \\begin{aligned} & {{m}^{2}}+1\\ne 1 \\\\ & {{m}^{2}}+1\\ne 2 \\\\ & \\left( {{m}^{2}}+1 \\right).1+m=3 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{m}^{2}}\\ne 0 \\\\ & {{m}^{2}}\\ne 1 \\\\ & {{m}^{2}}+m-2=0 \\\\ \\end{aligned} \\right.\\,$ $\\Leftrightarrow \\left\\{ \\begin{aligned} & {{m}^{2}}\\ne 0 \\\\ & {{m}^{2}}\\ne 1 \\\\ & m^2+2m-m-2=0 \\\\ \\end{aligned} \\right.\\,$ $\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & m\\ne \\pm 1 \\\\ & \\left( m-1 \\right)\\left( m+2 \\right)=0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & m\\ne \\pm 1 \\\\ & \\left[ \\begin{aligned} & m=1 \\\\ & m=-2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.\\Leftrightarrow m=-2$ <br\/>V\u1eady $m =-2$ th\u00ec ba \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u00e3 cho \u0111\u1ed3ng quy<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-2$.<\/span><br\/><br\/><span class='basic_green'>L\u01b0u \u00fd: <\/span> H\u1ecdc sinh hay qu\u00ean \u0111i\u1ec1u ki\u1ec7n hai \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau l\u00e0 $\\,a\\ne a'$.<\/span>"}]}],"id_ques":149},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"G\u1ecdi $\\,\\alpha\\,$ v\u00e0 $\\,\\beta\\,$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 g\u00f3c t\u1ea1o b\u1edfi \u0111\u01b0\u1eddng th\u1eb3ng $\\,y=2x+1\\,$ v\u00e0 $\\,y=x+2$ v\u1edbi tr\u1ee5c $Ox$. Ta c\u00f3:","select":["A. $\\alpha < \\beta < 90^o$","B. $\\alpha > \\beta > 90^o$","C. $\\beta < \\alpha < 90^o$","D. $\\beta = \\alpha < 90^o$ "],"hint":"\u0110\u01b0\u1eddng th\u1eb3ng $y=ax + b\\, (a > 0)$ t\u1ea1o v\u1edbi tia $Ox$ m\u1ed9t g\u00f3c $\\,\\alpha\\,$ th\u00ec $\\,a= tg\\alpha$","explain":"<span class='basic_left'> Do $\\alpha$ l\u00e0 g\u00f3c t\u1ea1o b\u1edfi \u0111\u01b0\u1eddng th\u1eb3ng $y=2x+1$ v\u1edbi tr\u1ee5c $Ox$ <br\/> $\\Rightarrow tg \\, \\alpha=2 \\Rightarrow \\alpha \\approx 63^o43'$<br\/>Do $\\beta$ l\u00e0 g\u00f3c t\u1ea1o b\u1edfi \u0111\u01b0\u1eddng th\u1eb3ng $y=x+2$ v\u1edbi tr\u1ee5c $Ox$<br\/> $\\Rightarrow tg\\,\\beta =1 \\Rightarrow \\beta= 45^o$ <br\/>Suy ra: $\\,\\beta < \\alpha < 90^o$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":150}],"lesson":{"save":0,"level":3}}