Question

Tìm x:(x-2)(x^2+2x+5)+2(x-2)(x+2)-5(x-2)=0

Answer 1

\((x-2)(x^2+2x+5+2x+4-5)=0<=> (x-2)(x^2+4x+4)=0 <=> (x-2)(x+2)^2 = 0 <=> x = 2 ; x = -2\)

Answer 2

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\left(x+2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)