Question

Giải pt

x^2-4=2(x-2)(x+3)

Answer 1

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

\(\Leftrightarrow x^2-4=2\left(x^2+3x-2x-6\right)\)

\(\Leftrightarrow x^2-4=2x^2+2x-12\)

\(\Leftrightarrow x^2-2x^2-2x=-12+4\)

\(\Leftrightarrow-x^2-2x=-8\)

\(\Leftrightarrow-x^2-2x+8=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}-x-4=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-4;2\right\}\)

Answer 2

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

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(x+2-2x-6\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Answer 3