Đáp án:c) x=-1
d)x=15
Giải thích các bước giải:
c)đk: \(x\neq±\frac{1}{3}\)
\(\frac{12}{1-9x^{2}}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
⇒\(\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}-\frac{12}{(1-3x)(1+3x)}=0\)
⇔\(\frac{(1-3x)(1-3x)-(1+3x)(1+3x)-12}{(1-3x)(1+3x)}=0\)
⇔\(\frac{9x^{2}-6x+1-9x^{2}-6x-1-12}{(1-3x)(1+3x)}=0\)
⇔\(\frac{-12x-12}{(1-3x)(1+3x)}=0\)
⇔-12x-12=0⇔ x=-1
d)đk:\(x\neq0; x\neq±5\)
\(\frac{x+5}{x^{2}-5x}-\frac{x^{2}+25}{2x^{2}-50} =\frac{x-5}{2x^{2}+10x}\)
⇒\(\frac{x+5}{x(x-5)}-\frac{x^{2}+25}{2(x-5)(x+5)}-\frac{x-5}{2x(x+5)}=0\)
⇔\(\frac{2(x+5)(x+5)-x(x+25)-(x-5)(x+5)}{2x(x-5)(x+5)}=0\)
⇔\(\frac{2x^{2}+20x+50-x^{2}-25x-x^{2}+25}{2x(x-5)(x+5)}=0\)
⇔\(\frac{-5x+75}{2x(x-5)(x+5)}=0\)
⇔-5x+75=0⇔ x=15
