Question

Answer 1

Answer 2

a) Ta có: \(2x^2-3x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

b) Ta có: \(x\left(x+1\right)-\left(x-2\right)^2=0\)

\(\Leftrightarrow x^2+x-x^2+4x-4=0\)

\(\Leftrightarrow5x=4\)

hay \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(x+1\right)\left(x-2\right)-\left(x-3\right)^2=x+1\)

\(\Leftrightarrow x^2-x-2-x^2+6x-9-x-1=0\)

\(\Leftrightarrow4x-12=0\)

\(\Leftrightarrow4x=12\)

hay x=3