Đáp án:

Giải thích các bước giải:

 a) $(\frac{x^2+1}{x^2-9}-\frac{x}{x+3}+\frac{5}{3-x}):(\frac{2x+10}{x+3}-1)\\
=(\frac{x^2+1}{x^2-9}-\frac{x(x-3)}{x^2-9}-\frac{5(x+3)}{x^2-9}):(\frac{2x+10}{x+3}-\frac{x+3}{x+3})\\
=\frac{x^2+1-x^2+3x-5x-15}{x^2-9}:\frac{2x+10-x-3}{x+3}\\
=\frac{-2x-14}{x^2-9}.\frac{x+3}{x+7}\\
=\frac{-2x-14}{(x-3)(x+7)}\\
=\frac{-2(x+7)}{(x-3)(x+7)}\\
=\frac{-2}{x+3}$

b) $|x-1|=2\Leftrightarrow x-1=2 (x-1>0),x-1=-2 (x-1<0)\\
\Leftrightarrow x=3(x>1),x=-1(x<1)$

khi đó $P=\frac{-2}{x+3}
x=3 \Rightarrow P=\frac{-2}{3+3}=\frac{-1}{3}\\
x=-1 \Rightarrow P=\frac{-2}{-1+3}=-1$

c) $x^2-3x+2=0\\
\Leftrightarrow x^2-x-2x+2=0\\
\Leftrightarrow x(x-1)-2(x-1)=0\\
\Leftrightarrow (x-1)(x-2)=0\\
\Leftrightarrow x-1=0,x-2=0\\
\Leftrightarrow x=1,x=2$

Khi đó $P=\frac{-2}{x+3}
x=1 \Rightarrow P=\frac{-2}{1+3}=\frac{-1}{2}\\
x=2 \Rightarrow P=\frac{-2}{2+3}=\frac{-2}{5}$

d) $\frac{-2}{x+3}=\frac{x+5}{6}\\
\Leftrightarrow -12=(x+5)(x+3)\\
\Leftrightarrow -12=x^2+3x+5x+15\\
\Leftrightarrow x^2+8x+27=0\\$