Giải phương trình
a) \(\dfrac{12}{1-9x^2}\)=\(\dfrac{1-3x}{1+3x}\)-\(\dfrac{1+3x}{1-3x}\)
b) \(\dfrac{2}{x^2-4}\)-\(\dfrac{x-1}{x\left(x-2\right)}\)+\(\dfrac{x-4}{x\left(x+2\right)}\)=0
c)\(\dfrac{16}{x^2-16}\)+\(\dfrac{2}{x+4}\)-\(\dfrac{1}{x-4}\)=0
d)\(\dfrac{x}{x-3}\)-\(\dfrac{x}{x-5}\)=\(\dfrac{x}{x-4}\)-\(\dfrac{x}{x-6}\)
\(a)\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(x\ne\pm\dfrac{1}{3}\right)\\ \Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\\ \Leftrightarrow12=\left(1-3x\right)^2-\left(1+3x\right)^2\\ \Leftrightarrow1-6x+9x^2-\left(1+6x+9x^2\right)=12\\ \Leftrightarrow1-6x+9x^2-1-6x-9x^2=12\\ \Leftrightarrow-12x=12\\ \Leftrightarrow x=-1\left(tm\right)\)
\(b)\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\left(x\ne0;\pm2\right)\\ \Leftrightarrow\dfrac{2x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow2x-\left(x-1\right)\left(x+2\right)+\left(x-4\right)\left(x-2\right)=0\\ \Leftrightarrow2x-\left(x^2+2x-x-2\right)+\left(x^2-2x-4x+8\right)=0\\ \Leftrightarrow2x-x^2-x+2+x^2-6x+8=0\\ \Leftrightarrow-5x+10=0\\ \Leftrightarrow-5x=-10\\ \Leftrightarrow x=2\left(ktm\right)\)
c)
\(\dfrac{16}{x^2-16}+\dfrac{2}{x+4}-\dfrac{1}{x-4}=0\left(x\ne\pm4\right)\\ \Leftrightarrow\dfrac{16}{\left(x+4\right)\left(x-4\right)}+\dfrac{2\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}=0\\ \Leftrightarrow16+2\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow16+2x-8-x+4=0\\ \Leftrightarrow x+12=0\\ \Leftrightarrow x=-12\left(tm\right)\)
