\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8=0\)

\(\Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

                                    Vậy S = { 0, 6}

(8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0

=> 8(x + 2) - 5x(x + 2) + 4[x(x + 1) - 2(x + 1)] + 2(x² - 4) = 0

=> 8x + 16 - 5x² - 10x + 4(x² + x - 2x - 2) + 2x² - 8 = 0

=> 8x + 16 - 5x² - 10x + 4x² + 4x - 8x - 8 + 2x² - 8 = 0

=> (8x - 10x + 4x - 8x) + (16 - 8 - 8) + (-5x² + 4x² + 2x²)  = 0

=> 0 + x² = 0

=> x² = 0 => x = 0

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(-5x^2-2x+16+4\left(x^2-x-2\right)+2\left(x^2-4\right)=0\)

\(-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)

\(x^2-6x=0\)

\(x\left(x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

( 8 - 5x )( x + 2 ) + 4( x - 2 )( x + 1 ) + 2( x - 2 )( x + 2 ) = 0

<=> ( x + 2 )[ ( 8 - 5x ) + 2( x - 2 ) ] + 4( x² - x - 2 ) = 0

<=> ( x + 2 )( 8 - 5x + 2x - 4 ) + 4x² - 4x - 8 = 0

<=> ( x + 2 )( 4 - 3x ) + 4x² - 4x - 8 = 0

<=> 4x - 3x² + 8 - 6x + 4x² - 4x - 8 = 0

<=> x² - 6x = 0

<=> x( x - 6 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

2: \(3x\left(x-4\right)+2x-8=0\)

=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(3x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

3: 4x(x-3)+x²-9=0

=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(4x+x+3\right)=0\)

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

=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4: \(x\left(x-1\right)-x^2+3x=0\)

=>\(x^2-x-x^2+3x=0\)

=>2x=0

=>x=0

5: \(x\left(2x-1\right)-2x^2+5x=16\)

=>\(2x^2-x-2x^2+5x=16\)

=>4x=16

=>x=4

Lời giải:

1. $(x+2)-2=0$

$x+2=2$

$x=0$

2.

$(x+3)+1=7$

$x+3=7-1=6$

$x=6-3=3$

3.

$(3x-4)+4=12$

$3x-4+4=12$

$3x=12$

$x=12:3=4$

4.

$(5x+4)-1=13$

$5x+4=13+1=14$

$5x=14-4=10$

$x=10:5=2$

5.

$(4x-8)-3=5$

$4x-8=5+3=8$

$4x=8+8=16$

$x=16:4=4$

6.

$3+(x-5)=7$

$x-5=7-3=4$

$x=4+5=9$

7.

$8-(2x-4)=2$

$2x-4=8-2=6$

$2x=6+4=10$

$x=10:2=5$

8.

$7+(5x+2)=14$

$5x+2=14-7=7$

$5x=7-2=5$

$x=5:5=1$

9.

$5-(3x-11)=1$

$3x-11=5-1=4$

$3x=11+4=15$

$x=15:3=5$

10.

$16-(8x+2)=6$

$8x+2=16-6=10$

$8x=10-2=8$

$x=8:8=1$

 (8 - 5x) (x + 2) + 4(x - 2) (x + 1) + 2(x - 2) (x + 2) = 0

=>  (x + 2) [ (8 - 5x) + 4(x + 1) + 2(x - 2)] = 0

=> (x + 2) (8 - 5x + 4x + 4 + 2x - 4)  = 0

=> (x + 2) (x + 8) = 0

=> x + 2 = 0   hoặc      x + 8 = 0

=> x = -2       hoặc        x = -8

a) (x+2)(x+3)-(x-2)(x+5)=0

  \(x^2+3x+2x+6-x^2-5x+2x+10=0\) 

\(2x+16=0\) 

\(2x=-16\) 

\(x=-8\) 

Vậy......

b) (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0

  \(8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2+4x-4x-8=0\) 

  \(-6x+x^2=0\) 

 \(x\left(-6+x\right)=0\) 

=> x=0   hoặc  -6+x=0  <=>x=6

Vậy \(x\in\left\{0;6\right\}\)

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

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

\(\Leftrightarrow x^2+2x+3x+6-x^2+5x+2x+10=0\)

\(\Leftrightarrow12x+16=0\)

\(\Leftrightarrow12x=-16\)

\(\Leftrightarrow x=\frac{-4}{3}\)

Vậy...

Bài tui sai rồi 2x ghi nhầm 12x