Question

Giải phương trình :\(\frac{2}{x^2-25}-\frac{1}{x^2+5x}=\frac{4}{x\left(x-5\right)}\)

Answer 1

ĐKXĐ: x≠0; x≠5; x≠-5

MTC=x(x+5)(x-5)

Ta có: \(\frac{2}{x^2-25}-\frac{1}{x^2+5x}=\frac{4}{x\left(x-5\right)}\)

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

\(\Leftrightarrow\frac{2x}{x\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x-5\right)\cdot x\cdot\left(x+5\right)}-\frac{4\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow2x-\left(x-5\right)-4\left(x+5\right)=0\)

\(\Leftrightarrow2x-x+5-4x-20=0\)

\(\Leftrightarrow-3x-15=0\)

\(\Leftrightarrow-3x=15\)

hay x=-5(ktm)

Vậy: x∈∅