Question

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

Answer 1

ĐKXĐ: \(x\ne1;x\ne-3\) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=0\)\(\Rightarrow3x^2+9x-x-3-2x^2+2x-5x+5+4-x^2+x-3x+3=0=3x+9=0\Rightarrow3x=9\Rightarrow x=3\)(thỏa mãn)

Vậy phương trình có tập nghiệm \(S=\left\{3\right\}\)