\(\frac{x^2-49}{x-7}+x-2=x+7+x-2=2x+9\)

\(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right)\frac{x^2+6x}{2x-6}\)

\(=\left(\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\right)\frac{x\left(x+6\right)}{2\left(x-3\right)}\)

\(=\frac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}.\frac{x\left(x+6\right)}{2\left(x-3\right)}=\frac{6\left(2x-6\right)}{x\left(x-6\right)}.\frac{x}{2\left(x-3\right)}\)

\(=\frac{12x}{2x\left(x-6\right)}=\frac{6}{x-6}\)

b, Ta có : \(\left|x+1\right|=3\)

TH1 : \(x+1=3\Leftrightarrow x=2\)

TH2 : \(x+1=-3\Leftrightarrow x=-4\)

Xét TH1 thay x = 2 ta có : 

\(P=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}\)

Xét TH2 thay x = -4 ta có : 

\(P=\frac{-4}{-4+2}=\frac{-4}{-2}=2\)

c, chưa học =)) 

\(P=\frac{x}{x+2}+\frac{2}{x-2}=\frac{2x+4}{4-x^2}\)

\(=\frac{x^2-2x}{\left(x+2\right)\left(x-2\right)}+\frac{2x+4}{\left(x-2\right)\left(x+2\right)}+\frac{2x+4}{\left(2-x\right)\left(x+2\right)}\)

\(=\frac{x^2-2x+2x+4-2x-4}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=\frac{x}{x+2}\)

Xét 2 trường hợp 

TH1 : 

\(x+10\ge0\)   

\(x\ge-10\)   

\(\left(x+1\right)^2+|x+10|-x^2-12=0\)   

\(x^2+2x+1+x+10-x^2-12=0\)   

\(3x-1=0\)   

\(x=\frac{1}{3}\left(n\right)\)   

TH2 : 

\(x+10< 0\)   

\(x< -10\)   

\(\left(x+1\right)^2+|x+10|-x^2-12=0\)   

\(x^2+2x+1-\left(x+10\right)-x^2-12=0\)   

\(x^2+2x+1-x-10-x^2-12=0\)   

\(x-21=0\)   

\(x=21\left(l\right)\)   

Vậy \(x=\frac{1}{3}\)   là nghiệm của phương trình 

(x + 1)² + |x + 10| - x² - 12 = 0

=> x² + 2x + 1 + |x + 10| - x² - 12 = 0

=> |x + 10| = -2x + 11

ĐKXĐ : -2x + 11 \(\ge0\Rightarrow x\le5\)

Khi đó |x + 10| = -2x + 11

<=> \(\orbr{\begin{cases}x+10=-2x+11\\x+10=2x-11\end{cases}}\Rightarrow\orbr{\begin{cases}3x=1\\x=21\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\left(\text{loại}\right)\\x=21\left(tm\right)\end{cases}}\Rightarrow x=21\)

Vậy x = 21 là giá trị cần tìm