Question

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

Answer 1

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

\(\Leftrightarrow\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}=\frac{12}{\left(1-3x\right)\left(1+3x\right)}\)

\(\Leftrightarrow\frac{\left(1-3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}-\frac{\left(1+3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}=\frac{12}{\left(1-3x\right)\left(1+3x\right)}\)

\(\Leftrightarrow\left(1-3x\right)\left(1+3x\right)-\left(1+3x\right)\left(1-3x\right)=12\)

\(0=12\)

=> x vô nghiệm

Answer 2

ĐKXĐ: \(\left(1+3x\right)\left(1-3x\right)\ne0\)

\(\Leftrightarrow\hept{\begin{cases}1-3x\ne0\\1+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\frac{1}{3}\\x\ne\frac{-1}{3}\end{cases}}}\)