Question

Bài 1: Giả cs phương trình

a)\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-1\right)}{x^2-4}\)

b) |3x| = x+8

(mink đag cần gấp)

Answer 1

a, \(\frac{\left(x-2\right)}{x+2}-\frac{3}{\left(x-2\right)}=\frac{2\left(x-1\right)}{x^2-4}\)

\(< =>\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x-2}{\left(x+2\right)\left(x-2\right)}\)

\(< =>\left(x-2\right)^2-\left(3x+6\right)=2x-2\)

\(< =>x^2-4x+4-3x-6=2x-2\)

\(< =>x^2-9x=0\)

\(< =>x\left(x-9\right)=0\)

\(< =>\left[{}\begin{matrix}x=0\\x-9=0\Leftrightarrow x=9\end{matrix}\right.\)

Vậy .........

Answer 2

b. |3x| = x+8

Điều kiện: \(x+8>0\) hay \(x>-8\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=x+8\\3x=-x-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=8\\4x=-8\end{matrix}\right.\)

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

Vậy ..........