{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'> T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $D=(a+b)(a+1)(b+1)$ bi\u1ebft $a+b=3$ v\u00e0 $ab=-5.$ <br\/> Gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $D$ l\u00e0: <\/span>","select":["A. $5$","B. $-3$","C. $-5$ ","D. $3$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - Thay $a+b=3$ v\u00e0o $D.$ <br\/> - Khai tri\u1ec3n $(a+1)(b+1)$ r\u1ed3i thay $a+b=3;ab=-5$ v\u00e0o \u0111\u1ec3 t\u00ednh. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & D=(a+b)(a+1)(b+1) \\\\ & \\,\\,\\,\\,\\,\\,=3\\left( a+1 \\right)\\left( b+1 \\right)\\,\\,(do\\,\\,a+b=3) \\\\ & \\,\\,\\,\\,\\,\\,=3\\left( ab+a+b+1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,=3\\left( -5+3+1 \\right) \\,\\,(do \\,\\,a+b=3;ab=-5) \\\\ & \\,\\,\\,\\,\\,\\,=3.\\left( -1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,=-3 \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span>","column":4}]}],"id_ques":1291},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c <br\/> $E={{x}^{10}}-2009{{x}^{9}}-2009{{x}^{8}}-...-2009x-1$ bi\u1ebft $x=2010$ <br\/> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $E$ l\u00e0: <\/span>","select":["A. $0$","B. $-1$","C. $2009$ ","D. $2010$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - \u0110\u00e1nh gi\u00e1 v\u1edbi $x=2010$ th\u00ec $x-1=2009$ <br\/> - Do \u0111\u00f3 ta thay $2009$ b\u1eb1ng $x-1$ v\u00e0o bi\u1ec3u th\u1ee9c $E$, bi\u1ebfn \u0111\u1ed5i r\u1ed3i t\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $E$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3 $x=2010 \\Rightarrow x-1=2009$ <br\/> Thay $2009$ b\u1eb1ng $x-1$ v\u00e0o $E$ ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align} & E={{x}^{10}}-2009{{x}^{9}}-2009{{x}^{8}}-...-2009x-1 \\\\ & \\,\\,\\,\\,\\,={{x}^{10}}-\\left( x-1 \\right){{x}^{9}}-\\left( x-1 \\right){{x}^{8}}-...-\\left( x-1 \\right)x-1 \\\\ & \\,\\,\\,\\,\\,={{x}^{10}}-{{x}^{10}}+{{x}^{9}}-{{x}^{9}}+{{x}^{8}}-...-{{x}^{2}}+x-1 \\\\ & \\,\\,\\,\\,\\,=x-1 \\\\ & \\,\\,\\,\\,\\,=2009 \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":4}]}],"id_ques":1292},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ecm gi\u00e1 tr\u1ecb nguy\u00ean c\u1ee7a bi\u1ebfn $x$ \u0111\u1ec3 bi\u1ec3u th\u1ee9c $B=\\dfrac{8-x}{x-3}$ c\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t. <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Ph\u00e2n t\u00edch $B$ th\u00e0nh d\u1ea1ng $a+\\dfrac{b}{3-x}$, trong \u0111\u00f3 $a,b$ l\u00e0 c\u00e1c s\u1ed1 nguy\u00ean \u00e2m. <br\/> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $\\dfrac{b}{3-x}$ \u0111\u1ec3 $B$ c\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $B=\\dfrac{8-x}{x-3}=\\dfrac{-\\left( x-8 \\right)}{x-3}=\\dfrac{-\\left( x-3-5 \\right)}{x-3}=-1+\\dfrac{5}{x-3}=-1-\\dfrac{5}{3-x}$ <br\/> \u0110\u1ec3 $B$ nh\u1ecf nh\u1ea5t th\u00ec $\\dfrac{5}{3-x}$ ph\u1ea3i l\u00e0 s\u1ed1 nguy\u00ean d\u01b0\u01a1ng l\u1edbn nh\u1ea5t <br\/> $\\Rightarrow 3-x=1 \\Rightarrow x=2$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$<\/span><\/span>"}]}],"id_ques":1293},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-83"],["11"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","input_hint":["frac"],"ques":"Cho <br\/> $\\begin{align} & F\\left( x \\right)={{x}^{3}}-2ax+{{a}^{2}} \\\\ & G\\left( x \\right)={{x}^{4}}+\\left( 3a+1 \\right)x+{{a}^{2}} \\\\ \\end{align}$ <br\/> T\u00ecm $a$ sao cho $F(1)=G(3).$<br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $a=$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & F\\left( 1 \\right)=G\\left( 3 \\right) \\\\ & \\Rightarrow {{1}^{3}}-2a.1+{{a}^{2}}={{3}^{4}}+\\left( 3a+1 \\right).3+{{a}^{2}} \\\\ & \\Rightarrow 1-2a+{{a}^{2}}=81+9a+3+{{a}^{2}} \\\\ & \\Rightarrow 11a+83=0 \\\\ & \\Rightarrow a=\\dfrac{-83}{11} \\\\ \\end{align}$<\/span>"}]}],"id_ques":1294},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> X\u00e9t \u0111a th\u1ee9c $P(x)=ax^2+bx+c\\,(a\\ne 0)$ <br\/> N\u1ebfu $a+b+c=0$ th\u00ec $P(x)$ c\u00f3 hai nghi\u1ec7m l\u00e0 $1$ v\u00e0 $\\dfrac{c}{a}.$<\/span>","select":["\u0110\u00fang","Sai "],"hint":"Thay $x=1;x=\\dfrac{c}{a}$ v\u00e0o $P(x)$ \u0111\u1ec3 ki\u1ec3m tra xem ch\u00fang c\u00f3 l\u00e0 nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)$","explain":"<span class='basic_left'> Thay $x=1$ v\u00e0o \u0111a th\u1ee9c $P(x)$ ta \u0111\u01b0\u1ee3c: <br\/> $a.1^2+b.1+c=a+b+c=0$ (v\u00ec $a+b+c=0$ theo gi\u1ea3 thi\u1ebft) <br\/> Do \u0111\u00f3, $x=1$ l\u00e0 nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)$ <br\/> Thay $x=\\dfrac{c}{a}$ v\u00e0o \u0111a th\u1ee9c $P(x)$ ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align} a.{{\\left( \\dfrac{c}{a} \\right)}^{2}}+b.\\dfrac{c}{a}+c &=\\dfrac{{{c}^{2}}}{a}+\\dfrac{bc}{a}+c \\\\ & =\\dfrac{{{c}^{2}}+bc+ac}{a} \\\\ & =\\dfrac{c\\left( c+b+a \\right)}{a} \\\\ & =0 \\\\ \\end{align}$ <br\/> (v\u00ec $a+b+c=0$ theo gi\u1ea3 thi\u1ebft) <br\/> Do \u0111\u00f3, $x=\\dfrac{c}{a}$ l\u00e0 nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":1295},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1"],["-1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ecm nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $(2x^2-3x+1)+(3x^2+3x-6)$ <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $\\left[ \\begin{align} & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right. $ <\/span> ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Cho \u0111a th\u1ee9c b\u1eb1ng $0$ r\u1ed3i bi\u1ebfn \u0111\u1ed5i t\u00ecm nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c \u0111\u00f3. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{aligned} & \\left( 2{{x}^{2}}-3x+1 \\right)+\\left( 3{{x}^{2}}+3x-6 \\right)=0 \\\\ & \\Rightarrow \\left( 2{{x}^{2}}+3{{x}^{2}} \\right)+\\left( -3x+3x \\right)+\\left( 1-6 \\right)=0 \\\\ & \\Rightarrow 5{{x}^{2}}-5=0 \\\\ & \\Rightarrow 5{{x}^{2}}=5 \\\\ & \\Rightarrow {{x}^{2}}=1 \\\\ & \\Rightarrow \\left[ \\begin{aligned} & x=1 \\\\ & x=-1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1;-1$<\/span><\/span>"}]}],"id_ques":1296},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho $x^2+y^2=1$ <br\/> Gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $2x^4+3x^2y^2+y^4+y^2$ l\u00e0: <\/span>","select":["A. $1$","B. $2$","C. $3$ ","D. $4$"],"hint":"T\u00e1ch $3x^2y^2=2x^2y^2+x^2y^2$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - T\u00e1ch $3x^2y^2=2x^2y^2+x^2y^2$ <br\/> - Nh\u00f3m c\u00e1c h\u1ea1ng t\u1eed v\u1edbi nhau th\u00e0nh t\u1eebng c\u1eb7p <br\/> - V\u1edbi t\u01b0ng c\u1eb7p, ta ph\u00e2n t\u00edch \u0111\u1ec3 l\u00e0m xu\u1ea5t hi\u1ec7n $x^2+y^2$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & 2{{x}^{4}}+3{{x}^{2}}{{y}^{2}}+{{y}^{4}}+{{y}^{2}} \\\\ & =(2{{x}^{4}}+2{{x}^{2}}{{y}^{2}})+({{x}^{2}}{{y}^{2}}+{{y}^{4}})+{{y}^{2}} \\\\ & =2{{x}^{2}}\\left( {{x}^{2}}+{{y}^{2}} \\right)+{{y}^{2}}\\left( {{x}^{2}}+{{y}^{2}} \\right)+{{y}^{2}} \\\\ & =2{{x}^{2}}.1+{{y}^{2}}.1+{{y}^{2}}\\,\\,\\,\\left( do\\,\\,{{x}^{2}}+{{y}^{2}}=1 \\right) \\\\ & =2{{x}^{2}}+{{y}^{2}}+{{y}^{2}} \\\\ & =\\left( {{x}^{2}}+{{y}^{2}} \\right)+\\left( {{x}^{2}}+{{y}^{2}} \\right) \\\\ & =1+1 \\\\ & =2 \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span> <\/span>","column":4}]}],"id_ques":1297},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> \u0110a th\u1ee9c $A(x)$ c\u00f3 \u00edt nh\u1ea5t ba nghi\u1ec7m n\u1ebfu bi\u1ebft: <br\/> $(x^2-4)A(x)=x.A(x-3)$<\/span>","select":["\u0110\u00fang","Sai "],"hint":"","explain":"<span class='basic_left'> Khi $x=2$ th\u00ec $(2^2-4)A(2)=2.A(2-3) \\Rightarrow A(-1)=0$ <br\/> Do \u0111\u00f3, $x=-1$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $A(x)$ <br\/> Khi $x=-2$ th\u00ec $[(-2)^2-4]A(-2)=(-2).A(-2-3) \\Rightarrow A(-5)=0$ <br\/> Do \u0111\u00f3, $x=-2$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $A(x)$<br\/> Khi $x=0$ th\u00ec $(0^2-4)A(0)=0.A(0-3) \\Rightarrow A(0)=0$ <br\/> Do \u0111\u00f3, $x=0$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $A(x)$ <br\/> V\u1eady \u0111a th\u1ee9c $A(x)$ c\u00f3 \u00edt nh\u1ea5t ba nghi\u1ec7m n\u1ebfu $(x^2-4)A(x)=x.A(x-3)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":1298},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> S\u1ed1 nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $x^2-6x+10$ l\u00e0: <\/span>","select":["A. $1$","B. $2$","C. Kh\u00f4ng c\u00f3 nghi\u1ec7m"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - T\u00e1ch $-6x=-3x-3x;10=9+1$ <br\/> - Nh\u00f3m c\u00e1c h\u1ea1ng t\u1eed v\u1edbi nhau th\u00e0nh t\u1eebng c\u1eb7p <br\/> - Ph\u00e2n t\u00edch, nh\u1eadn \u0111\u1ecbnh \u0111\u01b0a ra s\u1ed1 nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & {{x}^{2}}-6x+10 \\\\ & ={{x}^{2}}-3x-3x+9+1 \\\\ & =\\left( {{x}^{2}}-3x \\right)-\\left( 3x-9 \\right)+1 \\\\ & =x\\left( x-3 \\right)-3\\left( x-3 \\right)+1 \\\\ & =\\left( x-3 \\right)\\left( x-3 \\right)+1 \\\\ & ={{\\left( x-3 \\right)}^{2}}+1 \\\\ & Do\\,\\,{{\\left( x-3 \\right)}^{2}}\\,\\ge 0\\,\\,\\forall x \\\\ & \\Rightarrow {{\\left( x-3 \\right)}^{2}}+1\\ge 1>0 \\\\ \\end{align}$ <br\/> V\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $x$ \u0111\u1ec3 gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $x^2-6x+10$ b\u1eb1ng $0$ <br\/> Do \u0111\u00f3, \u0111a th\u1ee9c $x^2-6x+10$ kh\u00f4ng c\u00f3 nghi\u1ec7m. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span> <\/span>","column":3}]}],"id_ques":1299},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'> T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $M=12x^2+20x+1,$ bi\u1ebft $3x^2+5x-2=0.$ <\/span>","select":["A. $M=9$ v\u00e0 $M=2$","B. $M=-5$","C. $M=0$","D. $M=9$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x$ t\u1eeb $3x^2+5x-2=0.$ <br\/> V\u1ebf tr\u00e1i: <br\/> - T\u00e1ch $5x=6x-x$ <br\/> - Nh\u00f3m th\u00e0nh t\u1eebng c\u1eb7p v\u1edbi nhau r\u1ed3i bi\u1ebfn \u0111\u1ed5i. <br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay gi\u00e1 tr\u1ecb $x$ v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o \u0111a th\u1ee9c $M$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{aligned} & 3{{x}^{2}}+5x-2=0 \\\\ & \\Rightarrow 3{{x}^{2}}+6x-x-2=0 \\\\ & \\Rightarrow \\left( 3{{x}^{2}}+6x \\right)-\\left( x+2 \\right)=0 \\\\ & \\Rightarrow 3x\\left( x+2 \\right)-\\left( x+2 \\right)=0 \\\\ & \\Rightarrow \\left( x+2 \\right)\\left( 3x-1 \\right)=0 \\\\ & \\Rightarrow \\left[ \\begin{aligned} & x+2=0 \\\\ & 3x-1=0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left[ \\begin{aligned} & x=-2 \\\\ & x=\\dfrac{1}{3} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> Thay $x=-2$ v\u00e0o $M$ ta \u0111\u01b0\u1ee3c: <br\/> $12.(-2)^2+20.(-2)+1=9$ <br\/> Thay $x=\\dfrac{1}{3}$ v\u00e0o $M$ ta \u0111\u01b0\u1ee3c: <br\/> $12.{{\\left( \\dfrac{1}{3} \\right)}^{2}}+20.\\dfrac{1}{3}+1=9$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span> <br\/> <b> C\u00e1ch kh\u00e1c<\/b> <br\/> Ta c\u00f3 th\u1ec3 t\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $M$ nh\u01b0 sau: <br\/> $\\begin{align} & 12{{x}^{2}}+20x+1 \\\\ & =\\left( 12{{x}^{2}}+20x-8 \\right)+9 \\\\ & =4\\left( 3{{x}^{2}}+5x-2 \\right)+9 \\\\ & =4.0+9\\,\\,\\,\\left( do\\,\\,\\,3{{x}^{2}}+5x-2=0 \\right) \\\\ & =9 \\\\ \\end{align}$ <\/span>","column":2}]}],"id_ques":1300}],"lesson":{"save":0,"level":3}}