Question

[Nguyễn Đỗ Thục Quyên]

Giaỉ phương trình sau ;

4/x^2+2x-3 = 2x-5/x+3 - 2x/x-1

Answer 1

[Nguyễn Lê Phước Thịnh]

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

\(\Leftrightarrow\dfrac{\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{4}{\left(x+3\right)\left(x-1\right)}\)

Suy ra: \(2x^2-2x-5x+5-2x^2-6x=4\)

\(\Leftrightarrow13x=-1\)

hay \(x=-\dfrac{1}{13}\)