=>-7x^2+2+7x^2+14x=8

=>14x=6

=>x=3/7

`1)<=> -4x-3 + 5x+ 2 =0`

`<=> 5x-4x = -2+3`

`<=> x =1`

`2)<=> -5x +2-3x+6 =4`

`<=> -5x-3x = 4-6-2`

`<=> -8x=-4`

`<=> x=1/2`

`3) <=> -7x^2 +2 +7x^2 +14x =8`

`<=> 14x +2 =8`

`<=> 14x = 6`

`<=> x=3/7`

\(x^3-6x^2+12x-8=-8\\ \Leftrightarrow\left(x-2\right)^3=\left(-2\right)^3\\ \Leftrightarrow x-2=-2\\ \Leftrightarrow x=0\)

Vậy x = 0 là nghiệm của pt

\(x^3+6x^2+12x+8=0\)

\(\Rightarrow x^3+8+6x^2+12x=0\)

\(\Rightarrow x^3+2^3+3.2.x^2+3.2^2.x=0\)

\(\Rightarrow\left(x+2\right)^3=0\)

\(\Rightarrow x+2=0\Rightarrow x=-2\)

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

a) \(x\left(x-2\right)-7x+14=0\)

\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)

b) \(x^2\left(x-3\right)+12-4x=0\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)

c) \(x^2+12x-13=0\)

\(\Leftrightarrow\left(x^2-x\right)+\left(13x-13\right)=0\)

\(\Leftrightarrow x\left(x-1\right)+13\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)

d) \(4x^2-4x=8\)

\(\Leftrightarrow x^2-x-2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

e) \(x^2-6x=1\)

\(\Leftrightarrow\left(x-3\right)^2=10\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)

a) x( x - 2 ) - 7x + 14 = 0

<=> x( x - 2 ) - 7( x - 2 ) = 0

<=> ( x - 2 )( x - 7 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)

b) x²( x - 3 ) + 12 - 4x = 0

<=> x²( x - 3 ) - 4( x - 3 ) = 0

<=> ( x - 3 )( x² - 4 ) = 0

<=> \(\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)

c) x² + 12x - 13 = 0

<=> x² - x + 13x - 13 = 0

<=> x( x - 1 ) + 13( x - 1 ) = 0

<=> ( x - 1 )( x + 13 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\x+13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)

d) 4x² - 4x = 8

<=> 4( x² - x ) = 8

<=> x² - x = 2

<=> x² - x - 2 = 0

<=> x² + x - 2x - 2 = 0

<=> x( x + 1 ) - 2( x + 1 ) = 0

<=> ( x + 1 )( x - 2 ) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

e) x² - 6x = 1

<=> x² - 6x + 9 = 1 + 9

<=> ( x - 3 )² = 10

<=> ( x - 3 )² = ( ±√10 )²

<=> \(\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)