Question

Giair phương trình: \(\frac{x}{x^2-x-2}+\frac{3x}{x^2+x-2}=0\)

Answer 1

\(\dfrac{x}{{{x^2} - x + 2}} + \dfrac{{3x}}{{{x^2} + x - 2}} = 0\)

Điều kiện: \(x \ne -2;x \ne 1\)

\( \Leftrightarrow x\left( {{x^2} + x - 2} \right) + 3x\left( {{x^2} - x + 2} \right) = 0\\ \Leftrightarrow {x^3} + {x^2} - 2x + 3{x^3} - 3{x^2} + 6x = 0\\ \Leftrightarrow 4{x^3} - 2{x^2} + 4x = 0\\ \Leftrightarrow 2x\left( {2{x^2} - x + 2} \right) = 0\\ \Leftrightarrow x\left( {2{x^2} - x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0 \text{(nhận)}\\ 2{x^2} - x + 2 = 0\left( {VN} \right) \end{array} \right. \)