Question

Giải pt:

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

Answer 1

ĐKXĐ: \(x\ne\pm\dfrac{1}{3}\)

\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\\ \Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\\ \Rightarrow\left(1-3x\right)^2-\left(1+3x\right)^2=12\\ \Leftrightarrow\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\\ \Leftrightarrow-6x.2=12\\ \Leftrightarrow-6x=6\\ \Leftrightarrow x=-1\left(tmdk\right)\)