Question

1. (2x²-3x-4)²-(x²-x)²=0

2. (3x²+2x+4)²-(x²-4)²=0

Answer 1

Giúp mik vs ạ! Cần gấp

Answer 2

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

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

\(\Leftrightarrow\left[\left(x-1\right)^2-5\right]\left(x-2\right)\left(3x+2\right)=0\)

hay \(x\in\left\{\sqrt{5}+1;-\sqrt{5}+1;2;-\dfrac{2}{3}\right\}\)

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

\(\Leftrightarrow\left(2x^2+2x+8\right)\left(4x^2+2x\right)=0\)

hay \(x\in\left\{0;-\dfrac{1}{2}\right\}\)