{"segment":[{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho hai h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t <br\/>$y=\\left( m -\\dfrac{2}{3} \\right)x +3$ (1) v\u00e0 $y=(2-m)x+n - 1$(2)<br\/>\u0110\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 (1) v\u00e0 (2) l\u00e0 hai \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi nhau khi","select":["A. $m=\\dfrac{4}{3};\\,n\\ne 4$","B. $m\\ne \\dfrac{4}{3};\\,n\\ne 4$","C. $m=\\dfrac{4}{3};\\,n= 4$","D. $m\\ne\\dfrac{4}{3};\\,n= 4$ "],"hint":"Hai \u0111\u01b0\u1eddng th\u1eb3ng<br\/>$y=ax+b \\,\\,(a\\ne 0)$ v\u00e0 $y=a'x+b' \\,(a'\\ne 0)$ song song v\u1edbi nhau $\\Leftrightarrow\\, a=a';\\, b\\ne b'$","explain":"<span class='basic_left'> \u0110\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 (1) v\u00e0 (2) l\u00e0 hai \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi nhau khi: <br\/>$\\left\\{ \\begin{aligned} & m-\\dfrac{2}{3}=2-m \\\\ & 3\\ne n-1 \\\\ \\end{aligned} \\right.\\,$$\\,\\Leftrightarrow \\left\\{ \\begin{aligned} & 2m=\\dfrac{8}{3} \\\\ & n\\ne 4 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & m=\\dfrac{4}{3} \\\\ & n\\ne 4 \\\\ \\end{aligned} \\right.$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":171},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"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 $y=4x+3$ v\u00e0 $y=5x-2$ v\u1edbi tr\u1ee5c $Ox$. Ta c\u00f3:","select":["A. $\\alpha < \\beta < 90^0$","B. $\\alpha > \\beta > 90^0$","C. $\\beta < \\alpha < 90^0$","D. $\\beta = \\alpha < 90^0$ "],"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 $y=4x+3$ v\u1edbi tr\u1ee5c $Ox$. <br\/> $\\Rightarrow 4=tg \\, \\alpha \\Rightarrow \\alpha \\approx 75^057'$<br\/> Do $\\beta$ l\u00e0 g\u00f3c t\u1ea1o b\u1edfi \u0111\u01b0\u1eddng th\u1eb3ng $y=5x-2$ v\u1edbi tr\u1ee5c $Ox$. <br\/> $\\Rightarrow 5=tg\\,\\beta \\Rightarrow \\beta\\approx 78^041'$ <br\/>Suy ra $\\alpha < \\beta < 90^0$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":172},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"\u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 $y=|x|$ l\u00e0:","select":["A. <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/D924_TB4.png' \/><\/center>","B. <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/D924_TB5.png' \/><\/center>"],"hint":"\u00c1p d\u1ee5ng \u0111\u1ecbnh ngh\u0129a gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i, v\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00e3 cho tr\u00ean t\u1eebng kho\u1ea3ng.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: \u00c1p d\u1ee5ng \u0111\u1ecbnh ngh\u0129a gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i \u0111\u1ec3 vi\u1ebft l\u1ea1i h\u00e0m s\u1ed1 \u0111\u00e3 cho<br\/>B\u01b0\u1edbc 2: 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 $x$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3 $y=\\left\\{ \\begin{align} & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{x}\\ge \\text{0} \\\\ & -x\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{x}< \\text{0} \\\\\\end{align} \\right.$<br\/>V\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng $y = x$ v\u00e0 $y =-x$ tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. V\u00ec $|x|\\ge 0$ n\u00ean ta x\u00f3a ph\u1ea7n ph\u00eda d\u01b0\u1edbi tr\u1ee5c $Ox$ . <br\/> \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 l\u00e0 \u0111\u01b0\u1eddng g\u1ea5p kh\u00fac li\u1ec1n n\u00e9t nh\u01b0 h\u00ecnh v\u1ebd<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/D924_TB5.png' \/><\/center><br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span> V\u1edbi $A$ l\u00e0 m\u1ed9t bi\u1ec3u th\u1ee9c, ta c\u00f3<br\/>$|A|=\\left\\{ \\begin{align} & A\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{A}\\ge \\text{0} \\\\ & -A\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,\\text{A}< \\text{0} \\\\\\end{align} \\right.$ <\/span>","column":2}]}],"id_ques":173},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $\\dfrac{\\sqrt{2}}{2}$","B. $\\dfrac{2\\sqrt{2}}{2}$","C. $\\dfrac{3\\sqrt{2}}{2}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/10.jpg' \/><\/center>Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\,\\,\\,y=-x+3\\,\\,$. <br\/>Kho\u1ea3ng c\u00e1ch t\u1eeb g\u1ed1c t\u1ecda \u0111\u1ed9 $O$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ l\u00e0 ?","hint":" H\u1ea1 $OH$ vu\u00f4ng g\u00f3c v\u1edbi $(d)$. D\u00f9ng h\u1ec7 th\u1ee9c gi\u1eefa \u0111\u01b0\u1eddng cao v\u00e0 c\u1ea1nh trong tam gi\u00e1c vu\u00f4ng \u0111\u1ec3 t\u00ednh kho\u1ea3ng c\u00e1ch $OH.$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1:X\u00e1c \u0111\u1ecbnh giao \u0111i\u1ec3m $A, B$ c\u1ee7a \u0111\u1ed3 th\u1ecb v\u1edbi hai tr\u1ee5c t\u1ecda \u0111\u1ed9. <br\/>B\u01b0\u1edbc 2: H\u1ea1 $OH$ vu\u00f4ng g\u00f3c v\u1edbi $AB$. D\u00f9ng h\u1ec7 th\u1ee9c gi\u1eefa c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao trong tam gi\u00e1c $OAB$ \u0111\u1ec3 t\u00ednh $OH$.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/D924_TB3.png' \/><\/center><br\/>Ta c\u00f3 $\\left( d \\right)\\cap \\text{Ox}=A(3;0)$; $\\left( d \\right)\\cap Oy=B\\left( 0;3 \\right).$<br\/> H\u1ea1 $OH\\bot AB,$ $H$ thu\u1ed9c $AB.$<br\/>D\u00f9ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng $AOB,$ ta c\u00f3: <br\/>$\\begin{aligned}\\dfrac{1}{O{{H}^{2}}}&=\\dfrac{1}{O{{A}^{2}}}+\\dfrac{1}{O{{B}^{2}}}\\\\&=\\dfrac{1}{{{3}^{2}}}+\\dfrac{1}{{{3}^{2}}}=\\dfrac{2}{9}\\\\&\\Rightarrow OH=\\dfrac{3\\sqrt{2}}{2}\\\\ \\end{aligned}$ <br\/>Kho\u1ea3ng c\u00e1ch t\u1eeb $O$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $\\left( d \\right)$ l\u00e0 $\\dfrac{3\\sqrt{2}}{2}$<br\/><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span> <br\/>\u0110\u1ec3 gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp d\u1ea1ng t\u00ecm kho\u1ea3ng c\u00e1ch $h$ t\u1eeb g\u1ed1c t\u1ecda \u0111\u1ed9 \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng, ta th\u01b0\u1eddng s\u1eed d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng:<br\/> $\\dfrac{1}{{h}^{2}}=\\dfrac{1}{{b}^{2}}+\\dfrac{1}{{c}^{2}}$ v\u1edbi $b, c$ l\u00e0 c\u00e1c c\u1ea1nh g\u00f3c vu\u00f4ng.<span>"}]}],"id_ques":174},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"H\u00e0m s\u1ed1 $y=f\\left( x \\right)\\,$$=(\\sqrt{12-2\\sqrt{35}}-\\sqrt{7})x\\,$$+\\sqrt{5}$ lu\u00f4n \u0111\u1ed3ng bi\u1ebfn tr\u00ean t\u1eadp h\u1ee3p s\u1ed1 th\u1ef1c $\\mathbb{R}$.","select":["A. \u0110\u00fang","B. Sai"],"hint":"Tr\u00ean t\u1eadp h\u1ee3p s\u1ed1 th\u1ef1c $\\mathbb{R}$, h\u00e0m s\u1ed1 $y= ax + b$ \u0111\u1ed3ng bi\u1ebfn khi $a > 0$ v\u00e0 ngh\u1ecbch bi\u1ebfn khi $a < 0$.","explain":"<span class='basic_left'> H\u00e0m s\u1ed1 $y=f\\left( x \\right)\\,$$=(\\sqrt{12-2\\sqrt{35}}-\\sqrt{7})x\\,$$+\\sqrt{5}$ c\u00f3 d\u1ea1ng $y= ax + b$<br\/>V\u1edbi $a=\\sqrt{12-2\\sqrt{35}}-\\sqrt{7}\\\\=\\sqrt{\\sqrt{{{7}^{2}}}-2.\\sqrt{7}.\\sqrt{5}+\\sqrt{{{5}^{2}}}}-\\sqrt{7}\\\\ =\\sqrt{{{\\left( \\sqrt{7}-\\sqrt{5} \\right)}^{2}}}-\\sqrt{7}\\\\=\\sqrt{7}-\\sqrt{5}-\\sqrt{7}\\\\ =-\\sqrt{5} < 0$ <br\/>Suy ra h\u00e0m s\u1ed1 lu\u00f4n ngh\u1ecbch bi\u1ebfn tr\u00ean t\u1eadp s\u1ed1 th\u1ef1c $\\mathbb{R}$. <br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai<\/span><\/span>","column":2}]}],"id_ques":175},{"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":"Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$ cho ba \u0111i\u1ec3m $ A (0; 1);\\, B\\left(-\\dfrac{1}{2};0\\right)$ v\u00e0 $C(-3;m)$.<br\/>\u0110\u1ec3 ba \u0111i\u1ec3m $A, B, C$ th\u1eb3ng h\u00e0ng th\u00ec $m=$_input_","hint":"$A,B,C$ th\u1eb3ng h\u00e0ng khi v\u00e0 ch\u1ec9 khi \u0111i\u1ec3m $C$ thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $AB$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$.<br\/>B\u01b0\u1edbc 2: Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $C$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $AB$ \u0111\u1ec3 t\u00ecm $m$. <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(0;1) \\in AB \\Rightarrow a.0 + b= 1 \\Leftrightarrow b=1$<br\/>$B\\left(-\\dfrac{1}{2};0\\right) \\in AB \\Rightarrow a.\\left(-\\dfrac{1}{2}\\right)+b=0\\Leftrightarrow \\dfrac{-a}{2}=-1 \\Leftrightarrow a= 2$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ l\u00e0: $y= 2x + 1$<br\/>\u0110\u1ec3 $A, B, C$ th\u1eb3ng h\u00e0ng $\\Leftrightarrow C \\in AB$<br\/>$\\Leftrightarrow 2.(-3)+1 = m \\Leftrightarrow m=-5$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-5$.<\/span><\/span>"}]}],"id_ques":176},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho ba \u0111\u01b0\u1eddng th\u1eb3ng $y=x+7$; $y=2x+3$ v\u00e0 $y=mx-1$<br\/>\u0110\u1ec3 ba \u0111\u01b0\u1eddng th\u1eb3ng tr\u00ean \u0111\u1ed3ng quy th\u00ec $m=$ _input_","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: T\u00ecm giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng $y=x+7$; $y=2x+3$<br\/>B\u01b0\u1edbc 2: Thay t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $y=mx-1$. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> \u0110\u01b0\u1eddng th\u1eb3ng $y=x+7$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $y=2x+3$ t\u1ea1i $M$. <br\/> Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m $M$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$x + 7 = 2x + 3 \\,$$\\Rightarrow x = 4 \\Rightarrow y=11$<br\/>Suy ra $M (4; 11)$<br\/>\u0110\u1ec3 ba \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u1ed3ng quy th\u00ec \u0111\u01b0\u1eddng th\u1eb3ng $y=mx- 1$ \u0111i qua $M.$<br\/> Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $M$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $AB,$ ta c\u00f3: <br\/>$ m.4 - 1 =11 \\Leftrightarrow m = 3$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $3$.<\/span><\/span>"}]}],"id_ques":177},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"\u0110\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\,y= -\\dfrac{1}{3}x + 5$ v\u00e0 \u0111i qua \u0111i\u1ec3m $B (1; -4)$ c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: ","select":["A. $y= \\dfrac{5}{3}x+1$","B. $y=\\dfrac{1}{3}x -7$","C. $y= -3x + 1$","D. $y= 3x - 7$ "],"explain":"<span class='basic_left'> Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng c\u1ea7n t\u00ecm c\u00f3 d\u1ea1ng $(d'):\\,\\,y=ax + b$<br\/>$(d) \\bot (d') \\Rightarrow a.\\left(-\\dfrac{1}{3}\\right)= -1 \\Leftrightarrow a= 3$ <br\/>M\u00e0 $B (1; -4) \\,\\in (d')$ n\u00ean thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $B$ v\u00e0o $(d')$, ta \u0111\u01b0\u1ee3c: <br\/>$ 1.a + b= -4\\,$$ \\Rightarrow 3+ b= -4 \\Leftrightarrow b= -7$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $(d')$ c\u00f3 d\u1ea1ng $ y = 3x-7$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":2}]}],"id_ques":178},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-3"],["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/1.png' \/><\/center>Cho h\u00e0m s\u1ed1 $y=mx+3m +2$. T\u00ecm \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh $I$ m\u00e0 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 lu\u00f4n \u0111i qua v\u1edbi m\u1ecdi $m$.<br\/>T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $I$ l\u00e0 (_input_ ; _input_)","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: G\u1ecdi $I(x_{o};y_{o})$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh thu\u1ed9c $(d)$. Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m I v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ <br\/>B\u01b0\u1edbc 2: \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $a.m + b= 0$ v\u1edbi m\u1ecdi $m$.<br\/>B\u01b0\u1edbc 3: T\u00ecm $x_{o},y_{o}$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n \u0111\u00fang v\u1edbi m\u1ecdi $m.$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>X\u00e9t \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=mx+3m +2\\,\\,\\, (d)$<br\/>G\u1ecdi $I \\left( {{x}_{o}};{{y}_{o}} \\right)$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh c\u1ee7a $(d)$<br\/>$\\Rightarrow I\\left( {{x}_{o}};{{y}_{o}} \\right)\\in \\left( d \\right)\\,\\,\\,\\forall m$ <br\/>$\\begin{aligned} & {{y}_{o}}=m{{x}_{o}}+3m+2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\forall m \\\\ & \\Leftrightarrow m\\left( {{x}_{o}}+3 \\right)+2-{{y}_{o}}=0\\,\\,\\,\\forall m \\\\ \\end{aligned}$ <br\/>$ \\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{o}}+3=0 \\\\ & 2-{{y}_{o}}=0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{o}}=-3 \\\\ & {{y}_{o}}=2 \\\\ \\end{aligned} \\right.$ <br\/>$\\Rightarrow I\\left( -3;2 \\right)$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh $\\in \\left( d \\right)$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-3;2 $<\/span><\/span>"}]}],"id_ques":179},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/10.jpg' \/><\/center>H\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m $A (1; 3)$ v\u00e0 $B (2; 4)$ l\u00e0: ","select":["A. $1$","B. $-1$","C. $2$","D. $-2$"],"hint":"Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ c\u00f3 d\u1ea1ng: $y = ax + b\\, (a\\ne 0)$<br\/>Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $A$ v\u00e0 $B$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ c\u00f3 d\u1ea1ng $y= ax + b\\, (a \\ne 0)$<br\/>$A\\left( 1;3 \\right)\\,\\,\\in AB$$\\Rightarrow 3=a.1+b\\Leftrightarrow a+ b=3 \\,\\,\\,\\,(1)$<br\/>$B\\left( 2;4 \\right)\\in AB$$\\Rightarrow 4=a.2+b\\Leftrightarrow b= 4-2a\\,\\,\\,(2)$ <br\/>Thay (2) v\u00e0o (1) ta \u0111\u01b0\u1ee3c: $a + 4-2a = 3 \\Rightarrow a=1$<br\/>Suy ra: $b=4-2.1=2$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ l\u00e0: $y= x+ 2$<br\/>H\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $AB$ l\u00e0 $1$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":4}]}],"id_ques":180},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 $y=\\sqrt{16-{{x}^{2}}}+\\dfrac{3}{x-3}$ l\u00e0:","select":["A. $ -4\\le x\\le 4$ v\u00e0 $x\\ne 3$","B. $x \\ge 4$","C. $x \\le -4$","D. $ -4\\le x\\le 4$"],"hint":"<br\/>+ $\\sqrt{A}$ c\u00f3 ngh\u0129a $\\Leftrightarrow A \\ge 0$<br\/>+ $\\dfrac{a}{b}$ c\u00f3 ngh\u0129a $\\Leftrightarrow b \\ne 0$ ","explain":"<span class='basic_left'> H\u00e0m s\u1ed1 \u0111\u00e3 cho x\u00e1c \u0111\u1ecbnh khi <br\/>$\\left\\{ \\begin{aligned} & 16-{{x}^{2}}\\ge 0 \\\\ & x-3\\ne 0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}^{2}}\\le 16 \\\\ & x\\ne 3 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left| x \\right|\\le 4 \\\\ & x\\ne 3 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & -4\\le x\\le 4 \\\\ & x\\ne 3 \\\\ \\end{aligned} \\right.$ <br\/>V\u1eady \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 $ -4\\le x\\le 4$ v\u00e0 $x\\ne 3$. <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 A. <\/span><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span> $|x| \\le a \\Leftrightarrow -a \\le x \\le a$ v\u1edbi $a > 0$<\/span>","column":2}]}],"id_ques":181},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $M(\\dfrac{-6}{19};\\dfrac{-2}{19})$","B. $M(\\dfrac{-8}{19};\\dfrac{-2}{19})$","C. $M(\\dfrac{-10}{19};\\dfrac{-2}{19})$"],"ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d): \\,\\,y=5x +2$. T\u00ecm \u0111i\u1ec3m $M$ thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ sao cho \u0111i\u1ec3m \u0111\u00f3 c\u00f3 ho\u00e0nh \u0111\u1ed9 g\u1ea5p b\u1ed1n l\u1ea7n tung \u0111\u1ed9.<br\/><b>\u0110\u00e1p \u00e1n:<\/b> T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $M$ l\u00e0 (?;?)","hint":" $M(4x_{o},x_{o})\\in (d)$","explain":" <span class='basic_left'> Theo \u0111\u1ea7u b\u00e0i ta c\u00f3: $M\\left( 4{{x}_{o}};{{x}_{o}} \\right)$ <br\/>$M\\in \\left( d \\right)\\Rightarrow {{x}_{o}}=5.4{{x}_{o}}+2$$\\Leftrightarrow {{x}_{o}}=-\\dfrac{2}{19}$<br\/>T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $M\\left( -\\dfrac{8}{19}; -\\dfrac{2}{19}\\right)$"}]}],"id_ques":182},{"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":[[["2,25"]]],"list":[{"point":5,"width":60,"content":"","type_input":"","type_check":"","ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\, y= -2x + 3.\\,\\,(d)\\cap Ox =A,\\, (d)\\cap Oy=B$. <br\/><b> C\u00e2u 1:<\/b> Di\u1ec7n t\u00edch tam gi\u00e1c $AOB$ l\u00e0 _input_ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>(K\u1ebft qu\u1ea3 vi\u1ebft d\u01b0\u1edbi d\u1ea1ng s\u1ed1 th\u1eadp ph\u00e2n)","hint":"$S_{\\Delta}=\\dfrac{a.h}{2}$ ($a$ l\u00e0 c\u1ea1nh \u0111\u00e1y, $h$ l\u00e0 chi\u1ec1u cao)","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>D\u1ef1a v\u00e0o \u0111\u1ed3 th\u1ecb $y= -2x + 3$ \u0111\u1ec3 t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $AOB$.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\,(d)\\cap Ox =A,\\,\\, (d)\\cap Oy=B\\\\ \\Rightarrow A \\left( \\dfrac{3}{2};0 \\right),\\,\\,B(0; 3)$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/D924_TB1.png' \/><\/center><br\/>Theo h\u00ecnh v\u1ebd, ta c\u00f3 $OA= 1,5 ; OB= 3$.<br\/> Di\u1ec7n t\u00edch tam gi\u00e1c $AOB$ l\u00e0: $\\dfrac{OA.OB}{2}=\\dfrac{1,5.3}{2}=2,25$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch) <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2,25$.<\/span><\/span>"}]}],"id_ques":183},{"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"],["5"]]],"list":[{"point":5,"width":60,"content":"","type_input":"","type_check":"","ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\, y= -2x + 3. \\,\\,(d)\\cap Ox =A,\\, (d)\\cap Oy=B$<br\/><b> C\u00e2u 2:<\/b> Kho\u1ea3ng c\u00e1ch t\u1eeb \u0111i\u1ec3m $O$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ l\u00e0 $\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}}$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>H\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng. <br\/>$\\dfrac{1}{OH^2}=\\dfrac{1}{OA^2}+\\dfrac{1}{OB^2}$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> H\u1ea1 $\\,OH\\bot AB$ t\u1ea1i $H$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/D924_TB1.png' \/><\/center><br\/>Theo h\u00ecnh v\u1ebd, ta c\u00f3 $OA= 1,5 ; OB= 3$.<br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c $\\Delta AOB:$<br\/> $\\begin{aligned} & \\dfrac{1}{OH^2}=\\dfrac{1}{OA^2}+\\dfrac{1}{OB^2}\\\\ &\\dfrac{1}{OH^2}=\\dfrac{4}{9}+\\dfrac{1}{9}\\\\ &\\dfrac{1}{OH^2}=\\dfrac{5}{9}\\\\&\\Rightarrow OH= \\dfrac{3}{\\sqrt{5}}\\\\ \\end{aligned}$<br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $3$ v\u00e0 $5$.<\/span>"}]}],"id_ques":184},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{5}$","B. $2\\sqrt{5}$","C. $3\\sqrt{5}$"],"ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $(d):\\, y= -2x + 3. \\,(d)\\cap Ox =A,\\, (d)\\cap Oy=B$<br\/> Kho\u1ea3ng c\u00e1ch t\u1eeb \u0111i\u1ec3m $C(0; -2)$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $(d)$ l\u00e0 ?","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b> B\u01b0\u1edbc 1: <\/b> V\u1ebd h\u00ecnh. <br\/><b> B\u01b0\u1edbc 2:<\/b>Ch\u1ee9ng minh tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. <br\/><b> B\u01b0\u1edbc 3: <\/b> S\u1eed d\u1ee5ng t\u1ec9 l\u1ec7 c\u1ea1nh \u0111\u1ec3 t\u00ednh kho\u1ea3ng c\u00e1ch. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>H\u1ea1 $CK\\bot AB$ t\u1ea1i $K$ ; $BC=BO +OC = |3|+|2|=5$ <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/D924_TB2.png' \/><\/center><br\/>M\u00e0 $OH \\bot AB$ t\u1ea1i $H \\Rightarrow OH \/\/CK $<br\/>$\\Rightarrow \\Delta BOH \\backsim \\Delta BCK \\,$ (\u0110\u1ecbnh l\u00ed Talet)<br\/>$\\dfrac{OH}{CK}=\\dfrac{OB}{BC}\\,$$\\Leftrightarrow CK=\\dfrac{OH.BC}{OB}\\,$$=\\dfrac{\\dfrac{3}{\\sqrt{5}}.5}{3}=\\sqrt{5}$"}]}],"id_ques":185},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-2"],["-1"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/2.png' \/><\/center>Cho h\u00e0m s\u1ed1 $y=mx+2m -1$. T\u00ecm \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh $I$ m\u00e0 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 lu\u00f4n \u0111i qua v\u1edbi m\u1ecdi $m$. <br\/> T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $I$ l\u00e0 (_input_ ; _input_)","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: G\u1ecdi $I(x_{o};y_{o})$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh thu\u1ed9c $(d)$. Thay t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m $I$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng $(d)$.<br\/>B\u01b0\u1edbc 2: \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $a.m + b= 0$ v\u1edbi m\u1ecdi $m$.<br\/>B\u01b0\u1edbc 3: T\u00ecm $x_{o}; y_{o}$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n \u0111\u00fang v\u1edbi m\u1ecdi $m$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\,y=mx+2m -1\\,\\,\\, (d)$<br\/> G\u1ecdi $I \\left( {{x}_{o}};{{y}_{o}} \\right)$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh c\u1ee7a $(d)$ <br\/> $\\Rightarrow I\\left( {{x}_{o}};{{y}_{o}} \\right)\\in \\left( d \\right)\\,\\,\\,\\forall m$ <br\/> $\\begin{aligned} & {{y}_{o}}=m{{x}_{o}}+2m-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\forall m \\\\ & \\Leftrightarrow m\\left( {{x}_{o}}+2 \\right)-1-{{y}_{o}}=0\\,\\,\\,\\forall m \\\\ \\end{aligned}$ <br\/>$ \\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{o}}+2=0 \\\\ & -1-{{y}_{o}}=0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{o}}=-2 \\\\ & {{y}_{o}}=-1 \\\\ \\end{aligned} \\right.$ <br\/>$\\Rightarrow I\\left( -2;-1 \\right)$ l\u00e0 \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh $\\in \\left( d \\right)$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $\\left( -2;-1 \\right)$<\/span><\/span>"}]}],"id_ques":186},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"\u0110\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m $A (1; 5)$ v\u00e0 c\u1eaft tr\u1ee5c ho\u00e0nh t\u1ea1i \u0111i\u1ec3m $B$ c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng $2$ c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $y=-5x+10$","B. $y=5x - 10$","C. $y=-5x -10$ "],"hint":"Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ c\u00f3 d\u1ea1ng: $y = ax + b\\, (a\\ne 0)$.<br\/>Thay t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m $A$ v\u00e0 $B$ v\u00e0o \u0111\u01b0\u1eddng th\u1eb3ng. ","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ c\u00f3 d\u1ea1ng $y= ax + b\\, (a \\ne 0)$<br\/> \u0110\u01b0\u1eddng th\u1eb3ng c\u1eaft tr\u1ee5c ho\u00e0nh t\u1ea1i \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng $2 \\Rightarrow B (2; 0)$.<br\/>$A\\left( 1;5 \\right)\\,\\,\\in AB$$\\Rightarrow 5=a.1+b\\Leftrightarrow a+ b=5 \\,\\,\\,\\,(1)$<br\/>$B\\left( 2;0 \\right)\\in AB$$\\Rightarrow 0=a.2+b\\Leftrightarrow b= -2a\\,\\,\\,(2)$ <br\/>Thay (2) v\u00e0o (1) ta \u0111\u01b0\u1ee3c: $a - 2a = 5 \\Rightarrow a=-5$<br\/>Suy ra $b=-2.(-5)=10$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng $AB$ l\u00e0: $y= -5x +10$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":3}]}],"id_ques":187},{"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"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai12/lv2/img\/9.jpg' \/><\/center>Cho h\u00e0m s\u1ed1 $y=(k^2-5k+6)x -5$. T\u00ecm $k$ \u0111\u1ec3 h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn. <br\/>\u0110\u00e1p s\u1ed1: _input_ $< k <$ _input_","hint":"Tr\u00ean t\u1eadp h\u1ee3p s\u1ed1 th\u1ef1c $\\mathbb{R}$, h\u00e0m s\u1ed1 $y= ax + b$ \u0111\u1ed3ng bi\u1ebfn khi $a > 0$ v\u00e0 ngh\u1ecbch bi\u1ebfn khi $a < 0$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>Tr\u00ean t\u1eadp h\u1ee3p s\u1ed1 th\u1ef1c $\\mathbb{R}$, h\u00e0m s\u1ed1 $y= ax + b$ \u0111\u1ed3ng bi\u1ebfn khi $a > 0$ v\u00e0 ngh\u1ecbch bi\u1ebfn khi $a < 0$. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{aligned} & y=\\left( {{k}^{2}}-5k+6 \\right)x-5 \\\\ & \\Rightarrow y=(k^2-2k-3k+6)x-5 \\\\ & \\Rightarrow y=\\left( k-2 \\right)\\left( k-3 \\right)x-5 \\\\ \\end{aligned}$ <br\/>H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn $\\Leftrightarrow \\left( k-2 \\right)\\left( k-3 \\right)<0$ <br\/>$\\Leftrightarrow \\left[ \\begin{aligned} & \\left\\{ \\begin{aligned} & k-2>0 \\\\ & k-3<0 \\\\ \\end{aligned} \\right. \\\\ & \\left\\{ \\begin{aligned} & k-2<0 \\\\ & k-3>0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left[ \\begin{aligned} & \\left\\{ \\begin{aligned} & k>2 \\\\ & k< 3\\\\ \\end{aligned} \\right. \\\\ & \\left\\{ \\begin{aligned} & k<2 \\\\ & k>3 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow 2 < k < 3$ <br\/><span class='basic_pink'>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2$ v\u00e0 $3$.<\/span><\/span>"}]}],"id_ques":188},{"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","-1"],["-1","1"]]],"list":[{"point":5,"width":60,"content":"","type_input":"","type_check":"","ques":"Cho h\u00e0m s\u1ed1 $y=f\\left( x \\right)=\\left( {{m}^{2}}-1 \\right){{x}^{2}}+\\,$$\\left( m-5 \\right)x+3$. T\u00ecm $m$ \u0111\u1ec3 $y= f(x)$ l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t ngh\u1ecbch bi\u1ebfn. <br\/>\u0110\u00e1p \u00e1n: $m \\in$ {_input_;_input_}","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>H\u00e0m s\u1ed1 $y=ax^2+bx+c$ l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t ngh\u1ecbch bi\u1ebfn khi $a =0$ v\u00e0 $b < 0$. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> \u0110\u1ec3 $y=f(x)$ l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t ngh\u1ecbch bi\u1ebfn<br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{m}^{2}}-1=0 \\\\ & m-5<0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & {{m}^{2}}=1 \\\\ & m<5 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & m=\\pm 1 \\\\ & m<5 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow m=\\pm 1$ <br\/> <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$ v\u00e0 $-1$.<\/span><\/span>"}]}],"id_ques":189},{"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/bai12/lv2/img\/2.png' \/><\/center>Cho h\u00e0m s\u1ed1 $y=f\\left( x \\right)=\\left( m-\\sqrt{5} \\right)x+\\sqrt{5}+\\sqrt{3}$ <br\/>Cho $f\\left( 1 \\right)=2\\sqrt{3}.$ T\u00ecm $m$.","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $f\\left( 1 \\right)=2\\sqrt{3} $ \u0111\u1ec3 t\u00ecm $m$.","explain":"<span class='basic_left'>$\\begin{align}&\\,\\,\\,\\,\\, f\\left( 1 \\right)=2\\sqrt{3} \\\\& \\Leftrightarrow \\left( m-\\sqrt{5} \\right).1+\\sqrt{5}+\\sqrt{3}=2\\sqrt{3} \\\\ & \\Leftrightarrow m-\\sqrt{5} +\\sqrt{5}+\\sqrt{3}=2\\sqrt{3} \\\\ & \\Leftrightarrow m+\\sqrt{3}=2\\sqrt{3} \\\\ & \\Leftrightarrow m=\\sqrt{3} \\\\ \\end{align}$ <\/span> "}]}],"id_ques":190}],"lesson":{"save":0,"level":2}}