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1) \(\Rightarrow x^2+4x+4-x^2+1=9\)
\(\Rightarrow4x=4\Rightarrow x=1\)
2) \(\Rightarrow x\left(2x+7\right)+2\left(2x+7\right)=0\)
\(\Rightarrow\left(2x+7\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-2\end{matrix}\right.\)
3) \(\Rightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)
\(\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(-2x+1\right)\left(x+3\right)+\left(x+1\right)\left(2x-1\right)=14\)
\(\Leftrightarrow-2x^2-6x+x+3+2x^2-x+2x-1=14\)
\(\Leftrightarrow-4x=12\)
hay x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{2\left(x+2\right)\left(x-1\right)}{x+2}=2x-2\)
b: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=x^2-3x+1\)
c: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)
d: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
M = 4 x + 1 2 + 2 x + 1 2 − 8 x − 1 x + 1 − 12 x = 4 ( x 2 + 2 x + 1 ) + ( 4 x 2 + 4 x + 1 ) – 8 ( x 2 – 1 ) – 12 x = 4 x 2 + 8 x + 4 + 4 x 2 + 4 x + 1 – 8 x 2 + 8 – 12 x = 4 x 2 + 4 x 2 − 8 x 2 + 8 x + 4 x − 12 x + 4 + 1 + 8 = 13
N = 2 ( x – 1 ) 2 – 4 ( 3 + x ) 2 + 2 x ( x + 14 ) = 2 x 2 − 2 x + 1 − 4 9 + 6 x + x 2 + 2 x 2 + 28 x = 2 x 2 − 4 x + 2 − 36 − 24 x − 4 x 2 + 2 x 2 + 28 x = ( 2 x 2 + 2 x 2 – 4 x 2 ) + ( - 4 x – 24 x + 28 x ) + 2 – 36 = - 34
Suy ra M = 13, N = -34 ó 2M – N = 60
Đáp án cần chọn là: B
![](https://rs.olm.vn/images/avt/0.png?1311)
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-x\left(x+2\right)=6\)
\(\Leftrightarrow x^2-x^2-2x=6\)
<=> -2x = 6
<=> x = -3
\(3x\left(x-2\right)+2x\left(2-x\right)=x^2-8\)
\(\Leftrightarrow3x\left(x-2\right)-2x\left(x-2\right)=x^2-8\)
\(\Leftrightarrow\left(x-2\right)\left(3x-2x\right)=x^2-8\)
\(\Leftrightarrow\left(x-2\right)x=x^2-8\)
\(\Leftrightarrow x^2-2x=x^2-8\)
\(\Leftrightarrow2x=8\)
<=> x = 4
a/ \(x^2-x\left(x+2\right)=6\)
<=> \(x^2-x^2-2x=6\)
<=> \(-2x=6\)
<=> \(x=-3\)
b/ \(3x\left(x-2\right)+2x\left(2-x\right)=x^2-8\)
<=> \(3x^2-6x+4x-2x^2=x^2-8\)
<=> \(3x^2-2x-2x^2-x^2+8=0\)
<=> \(-2x+8=0\)
<=> \(-2x=-8\)
<=> \(x=4\)
c/ \(3\left(5x-1\right)-x\left(x+1\right)+x^2=14\)
<=> \(15x-3-x^2-x+x^2=14\)
<=> \(14x-3=14\)
<=> \(-3=14-14x\)
<=> \(14\left(1-x\right)=-3\)
<=> \(1-x=\frac{-3}{14}\)
<=> \(-x=\frac{-3}{14}-1\)
<=> \(x=\frac{3}{14}+1\)
<=> \(x=\frac{17}{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x(2x-7)-4x+14=0
=>x(2x-7)-2(2x-7)=0
=>(x-2)(2x-7)=0
=>x-2=0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
b, x(x-1)+2x-2=0
=>x(x-1)+2(x-1)=0
=>(x+2)(x-1)=0
=>x+2=0 hoặc x-1=0
=>x=-2 hoặc x=1
c, 2x^3+3x^2+2x+3=0
=>x2(2x+3)+2x+3=0
=>(x2+1)(2x+3)=0
=>x2+1=0 hoặc 2x+3=0
Vì x2+1>0 với mọi x ->vô nghiệm
=>2x+3=0 =>x=-3/2
d, x^3+6x^2+11x+6=0
=>x3+3x3+2x+3x2+3x3+6=0
=>x(x2+3x+2)+3(x2+3x+2)=0
=>(x2+3x+2)(x+3)=0
=>[x2+x+2x+2](x+3)=0
=>[x(x+1)+2(x+1)](x+3)=0
=>(x+1)(x+2)(x+3)=0
=>x+1=0 hoặc x+2=0 hoặc x+3=0
=>x=-1 hoặc x=-2 hoặc x=-3![](https://rs.olm.vn/images/avt/0.png?1311)
a) x(2x-7)-4x+14=0
=>x(2x-7)-2(2x-7)=0
=>(x-2)(2x-7)=0
=>x-2=0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
b, x(x-1)+2x-2=0
=>x(x-1)+2(x-1)=0
=>(x+2)(x-1)=0
=>x+2=0 hoặc x-1=0
=>x=-2 hoặc x=1
c, 2x^3+3x^2+2x+3=0
=>x2(2x+3)+2x+3=0
=>(x2+1)(2x+3)=0
=>x2+1=0 hoặc 2x+3=0
Vì x2+1>0 với mọi x ->vô nghiệm
=>2x+3=0 =>x=-3/2
d, x^3+6x^2+11x+6=0
=>x3+3x3+2x+3x2+3x3+6=0
=>x(x2+3x+2)+3(x2+3x+2)=0
=>(x2+3x+2)(x+3)=0
=>[x2+x+2x+2](x+3)=0
=>[x(x+1)+2(x+1)](x+3)=0
=>(x+1)(x+2)(x+3)=0
=>x+1=0 hoặc x+2=0 hoặc x+3=0
=>x=-1 hoặc x=-2 hoặc x=-3
\(\Leftrightarrow x^3+8-x^3+3x=14\)
=>3x=6
hay x=2
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