Dễ thế mà không làm được

<= 2x^2 - 6x = 3x^2 + 3x - x^2 + 2x + 5 

<=> 2x^2 - 6x = 2x^2 + 5x + 5

<=> -6x - 5x     = 5 

=> -11x             = 5 

=> x                  = -5/11 

Nhân ra rút gọn  

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

 20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1

 10x2−19x−33=010x2−19x−33=0

 (10x+11)(x−3)=0

chỉ bt lm con b thoy

..army,,,,,,,,,,

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

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

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

\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\frac{22}{5}\)

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

\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2\)

\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)

\(\Leftrightarrow20x^2-16x-33=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

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 - 10x - 2 = 3x² - 15x + 20
<-> 3x² - 3x² - 10x + 15x = 20 + 2
<-> 5x = 22
<-> x = 22/5

áp dụng các hằng đẳng thức thôi mà :)

a)\(x^2-2x+1=25\)

=>\(\left(x-1\right)^2=25\)

=>\(\orbr{\begin{cases}x-1=-5\\x-1=5\end{cases}}\)

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

=>\(3\left[\left(x-1\right)^2-x\left(x-5\right)\right]=1\)

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

=>\(3\left(3x+1\right)=1\)

=>\(3x+1=\frac{1}{3}\)

=>\(3x=\frac{-2}{3}\)

=>\(x=\frac{-2}{9}\)

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

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

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

=>\(\orbr{\begin{cases}5-2x-4=0\\5-2x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}}\)