\(a,ĐK:x\ne\pm5\\ b,C=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{16}{3}\\ Q=9x^2-42x+49=\left(3x-7\right)^2\\ Q=\left[3\left(-\dfrac{16}{3}\right)-7\right]^2=\left(-16-7\right)^2=529\)
a) ĐKXĐ: \(x\ne5,x\ne-5\)
b) \(C=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2}{x-5}=\dfrac{1}{x+5}\)
c) \(P=\dfrac{1}{x+5}=-3\Leftrightarrow-3x-15=1\Leftrightarrow x=-\dfrac{16}{3}\left(tm\right)\)
\(Q=9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-16-7\right)^2=23^2=529\)
