Question

Giải các phương trình sau:

a) \(\frac{2x-3}{x^2-9}-\frac{2x}{3-x}=\frac{-5}{x+3}\)

b) \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}=1\)

Answer 1

a/ ĐK: \(x\ne\pm3\)

\(\frac{2x-3}{x^2-9}+\frac{2x\left(x+3\right)}{x^2-9}+\frac{5\left(x-3\right)}{x^2-9}=0\)

\(\Leftrightarrow2x-3+2x^2+6x+5x-15=0\)

\(\Leftrightarrow2x^2+13x-18=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{-13+\sqrt{313}}{4}\\x=\frac{-13-\sqrt{313}}{4}\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ne1;-3\)

\(\frac{\left(3x-1\right)\left(x+3\right)}{x^2+2x-3}-\frac{\left(2x+5\right)\left(x-1\right)}{x^2+2x-3}+\frac{4}{x^2+2x-3}-\frac{x^2+2x-3}{x^2+2x-3}=0\)

\(\Leftrightarrow3x^2+8x-3-\left(2x^2+3x-5\right)+4-\left(x^2+2x-3\right)=0\)

\(\Leftrightarrow3x+9=0\)

\(\Rightarrow x=-3\) (ko thỏa mãn)

Vậy pt vô nghiệm