Question

`h) (4x^2-3x-18)^2 = (4x^2 + 3x)^2` 

Giải pt

Answer 1

\(\Leftrightarrow\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)=0\)

\(\Leftrightarrow\left(-6x-18\right)\left(8x^2-18\right)=0\)

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

hay \(x\in\left\{-3;\dfrac{3}{2};-\dfrac{3}{2}\right\}\)

Answer 2

<=> 4x²-3x-18=4x²+3x

<=>6x=-18

<=>x=-3