Question

Giải phương trình:

\(\dfrac{x}{x+2}=\dfrac{4x^2-x-4}{x^2-4}+\dfrac{3x-1}{2-x}\)

Answer 1

ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{x}{x+2}=\dfrac{4x^2-x-4}{x^2-4}+\dfrac{3x-1}{2-x}\)

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

\(\Rightarrow x^2-2x=4x^2-x-4-3x^2-5x+2\)

\(\Leftrightarrow4x=-2\)

\(\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

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