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

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) \(\Rightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)

\(\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)

\(x^2-x\left(x+2\right)=6\)

\(\Leftrightarrow x^2-x^2-2x=6\)

<=> -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/ \(3\left(5x-1\right)-x\left(x+1\right)+x^2=14\)

<=> \(15x-3-x^2-x+x^2=14\)

<=> \(14x-3=14\)

<=> \(-3=14-14x\)

<=> \(14\left(1-x\right)=-3\)

<=> \(1-x=\frac{-3}{14}\)

<=> \(-x=\frac{-3}{14}-1\)

<=> \(x=\frac{3}{14}+1\)

<=> \(x=\frac{17}{14}\)

\(\Leftrightarrow x^3+8-x^3+3x=14\)

=>3x=6

hay x=2

2

* là dấu nhân nha mn 

\(\left(2x+1\right)\left(y+2\right)=14\)

\(\Rightarrow\hept{\begin{cases}2x+1\\y+2\end{cases}}\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

mà \(2x+1\) là số lẻ  \(\Rightarrow\)   \(\left(2x+1\right)\in\left\{\pm1;\pm7\right\}\)

Ta có bảng:

Vậy \(\left(x,y\right)=\left(0;1\right),\left(-1;-3\right),\left(3;5\right);\left(-4;-9\right)\)

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

giúp mình với

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

tau cung bui ma chu mi giup tao roi cam on nhe

\(\left|x^2-2x-1\right|=14\)

\(\Rightarrow\orbr{\begin{cases}x^2-2x-1=14\\x^2-2x-1=-14\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x^2-2x=15\\x^2-2x=-13\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x\in\varnothing\end{cases}}}\)

\(\Rightarrow x=-3\)

\(ok\)