Question

Giải phương trình:

a) \(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)

b) \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

Answer 1

a) ĐKXĐ: \(x\notin\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)

Ta có: \(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)

\(\Leftrightarrow\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}\)

Suy ra: \(9x^2-6x+1-9x^2-6x-1=12\)

\(\Leftrightarrow-12x=12\)

hay x=-1(thỏa ĐK)

Vậy: S={-1}