Question

giải phương trình:

\(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)

Answer 1

Lời giải:

ĐKXĐ:.....

Ta có: \(\frac{2x}{3x^2-x+2}-\frac{7x}{3x^2+5x+2}=1\)

\(\Leftrightarrow \frac{1}{6}+\frac{2x}{3x^2-x+2}-7\left(\frac{x}{3x^2+5x+2}+\frac{1}{6}\right)=0\)

\(\Leftrightarrow \frac{3x^2+11x+2}{6(3x^2-x+2)}-\frac{7(3x^2+11x+2)}{6(3x^2+5x+2)}=0\)

\(\Leftrightarrow \frac{1}{6}(3x^2+11x+2)\left(\frac{1}{3x^2-x+2}-\frac{7}{3x^2+5x+2}\right)=0\)

TH1: \(3x^2+11x+2=0\)

\(\Leftrightarrow x=\frac{-11\pm \sqrt{97}}{6}\) (thỏa mãn)

TH2: \(\frac{1}{3x^2-x+2}-\frac{7}{3x^2+5x+2}=0\)

\(\Leftrightarrow \frac{2}{3x^2-x+2}-\frac{7}{3x^2+5x+2}=\frac{1}{3x^2-x+2}\)

\(\Leftrightarrow \frac{1}{x}=\frac{1}{3x^2-x+2}\)

\(\Leftrightarrow x=3x^2-x+2\)

\(\Leftrightarrow 3x^2-2x+2=0\)

\(\Leftrightarrow 2x^2+(x-1)^2+1=0\) (vô lý)

Do đó PT có nghiệm \(x=\frac{-11\pm \sqrt{97}}{6}\)