\(a.\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-4x-4=5\)

\(\left(-4x-6x\right)+\left(4-9\right)-4x-4=5\)

\(-10x-5-4x-4=5\)

\(-14x-9=5\)

\(-14x=14\Rightarrow x=-1\)

\(b.\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(17x-10=-44\)

\(17x=-34\Rightarrow x=-2\)

\(c.\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(25x^2+10x+1-\left(25x^2-9\right)=30\)

\(10x+10=30\)

\(10x=20\Rightarrow x=2\)

\(d.\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)

\(\left(x^2+6x+9\right)+\left(x^2-4\right)-2\left(x^2-2x+1\right)=7\)

\(2x^2+6x+5-2x^2+4x-2=7\)

\(10x+3=7\)

\(10x=4\Rightarrow x=\frac{4}{10}=\frac25\)

\(f.\left(3x-8\right)^2=0\)

\(3x-8=0\Rightarrow x=\frac83\)

\(e.6\left(x+1\right)^2-2\left(x+1\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)

\(6\left(x^2+2x+1\right)-2x-2+2\left(x^3-1\right)=0\)

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

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

\(\Rightarrow x\approx-0,23\)

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

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

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

                                    Vậy S = { 0, 6}

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

 (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

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

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

\(2x+16=0\)

\(2x=-16\)

\(x=-8\)

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

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

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

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

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

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

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

\(\Rightarrow2x=-16\Leftrightarrow x=-8\)

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

\(\Rightarrow8x+16-5x^2-10x+4\left(x^2-x+2\right)+2\left(x^2-4\right)=0\)

\(\Rightarrow8x+16x-5x^2-10x+4x^2-4x+8+2x^2-8=0\)

\(\Rightarrow x^2+10x=0\Rightarrow x\left(x+10\right)=0\Rightarrow x\in\left\{0;-10\right\}\)