Đáp án + Giải thích các bước giải:

a) `(2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)`

`2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20`

`3x² - 12x - 2 - 3x² + 17x = 20`

`5x = 22`

`x = \frac{22}{5}`

b)

`(8x - 3)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1)`

`24x² + 16x - 9x - 6 - (4x² + 16x + 7x + 28) = 10x² - 2x + 5x - 1`

`24x² + 7x - 6 - 4x² - 23x - 28 - 10x² - 3x + 1 = 0`

`10x² - 19x - 33 = 0`

`10x² + 11x - 30x - 33 = 0`

`(x-3)(10x + 11) = 0`

\(\left[ \begin{array}{l}x=3\\x=-\dfrac{11}{10}\end{array} \right.\) 

c)

`2x² + 3(x - 1)(x + 1) = 5x(x + 1)`

`2x² + 3x² - 3 = 5x² + 5x`

`5x² - 5x² - 3 = 5x`

`5x = -3`

`x = \frac{-3}{5}`

d)

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

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

`-5x² - 2x + 16 + 4x² - 4x - 8 + x² = 4`

`-6x + 8 = 4`

`6x = 4`

`x = \frac{2}{3}` 

e)` 4(x - 1)( x + 5) - (x +2)(x + 5) = 3(x - 1)(x + 2)`

`4(x² + 5x - x - 5) - (x² + 7x + 10) = 3(x² + x - 2)`

`4x² + 16x - 20 - x² - 7x - 10 - 3x² - 3x = -6``

`6x - 30 = -6`

`6x = 24`

`x = 4`

\(a)\)

\(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\)

\(\Leftrightarrow3x^2-12x-2-3x^2+17x-20=0\)

\(\Leftrightarrow5x-22=0\)

\(\Leftrightarrow x=\dfrac{22}{5}\)

\(b)\)

\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)-\left(2x+1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left(24x^2+16x-9x-6\right)-\left(4x^2+16x+7x+28\right)-\left(10x^2-2x+5x-1\right)=0\)

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28-10x^2+2x-5x+1=0\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Leftrightarrow10x^2+11x-30x-33=0\)

\(\Leftrightarrow\left(10x^2+11x\right)-\left(30x+33\right)=0\)

\(\Leftrightarrow x\left(10x+11\right)-3\left(10x+11\right)=0\)

\(\Leftrightarrow\left(10x+11\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}10x+11=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}10x=-11\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11}{10}\\x=3\end{matrix}\right.\)

Vậy \(x=\dfrac{-11}{10}\) hoặc \(x=3\)

\(c)\)

\(2x^2+3(x-1)(x+1)=5x(x+1)\)

 \(\Leftrightarrow2x^2+3\left(x^2+x-x-1\right)=5x^2+5x\)

\(\Leftrightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)

\(\Leftrightarrow2x^2+3x^2-3=5x^2+5x\)

\(\Leftrightarrow5x^2-3=5x^2+5x\)

\(\Leftrightarrow5x=-3\)

\(\Leftrightarrow x=\dfrac{-3}{5}\)

Vậy \(x=\dfrac{-3}{5}\)

\(d)\)

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

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

\(\Rightarrow16-5x^2-2x+4x^2-4x-8+2x^2-8=0\)

\(\Rightarrow x^2-6x=0\Rightarrow x.\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

\(e)\)

\(4(x - 1)( x + 5) - (x +2)(x + 5) = 3(x - 1)(x + 2)\)

\(\Leftrightarrow(4x-4)(x+5)-(x^2+5x+2x+10)=(3x-3)(x+2)\)

\(\Leftrightarrow4x^2+20x-4x-20-x^2-5x-2x-10=3x^2+6x-3x-6\)

\(\Leftrightarrow3x^2+9x-3x^2-6x+3x=-6+10+20\)

\(\Leftrightarrow6x=24\)

\(\Leftrightarrow x=4\)