Bài 6
\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow\dfrac{1}{x+5}=-3\Leftrightarrow-3\left(x+5\right)=1\Leftrightarrow x=-\dfrac{16}{3}\\ \Leftrightarrow Q=\left(3x-7\right)^2=\left[3\cdot\left(-\dfrac{16}{3}\right)-7\right]^2=529\)
Bài 7:
\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\\ b,P=4\Leftrightarrow4\left(x-3\right)=4\Leftrightarrow x=4\)
Điều kiện (x≠5, x≠-5)
\(P=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}+\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}-\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{x-5+2\left(x+5\right)-2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
Bài 8:
a. ĐKXĐ: $x\neq \pm 2$
\(A=\frac{x^2}{(x-2)(x+2)}-\frac{x(x+2)}{(x-2)(x+2)}+\frac{2(x-2)}{(x-2)(x+2)}=\frac{x^2-x(x+2)+2(x-2)}{(x-2)(x+2)}\)
\(=\frac{-4}{x^2-4}\)
b. Tại $x=1$ thì $A=\frac{-4}{1-4}=\frac{4}{3}$
