Question

Tìm x

A) (2x + 3)(x -4) + (x-5)(x-2)=(3x-5)(x-4)

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

Answer 1

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

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

\(\Leftrightarrow\left(2x^2-8x+3x-12\right)+\left(x^2-2x-5x+10\right)-\left(3x^2-12x-5x+20\right)=0\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10-3x^2+12x+5x-20=0\)

\(\Leftrightarrow5x-22=0\)

\(\Leftrightarrow5x=22\)

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

Vậy \(x=\dfrac{22}{5}\)