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

đkxđ: \(x\ne0;x\ne\pm6\)

MTC=x(x+6)(x-6)

\(=\left[\frac{x}{\left(x+6\right)\left(x-6\right)}+\frac{6-x}{x\left(x+6\right)}\right]\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)

\(=\left[\frac{x^2}{x\left(x^2-36\right)}-\frac{\left(x-6\right)^2}{x\left(x^2-36\right)}\right]\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)

\(=\frac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)

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

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

.....................

a)

\(S=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)

\(S=\left(\dfrac{x}{\left(x+6\right)\left(x-6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(S=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(S=\left(\dfrac{x^2-x^2+12x-36}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(S=\dfrac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(S=\dfrac{6}{x-6}-\dfrac{x}{x-6}\)

\(S=\dfrac{6-x}{x-6}=-1\)

b) Vì giá trị của biểu thức S không phụ thuộc vào giá trị của biến nên với mọi giá trị của x ta đều có giá trị của S là - 1.

Sửa đề: \(\dfrac{3}{x^2+6x+9}-\dfrac{3}{x^2-6x+9}+\dfrac{x^2+30x-27}{x^4-18x^2+81}\)

\(=\dfrac{3x^2-18x+27-3x^2-18x-27+x^2+30x-27}{\left(x+3\right)^2\cdot\left(x-3\right)^2}\)

\(=\dfrac{x^2-6x-27}{\left(x+3\right)^2\cdot\left(x-3\right)^2}=\dfrac{\left(x-9\right)\left(x+3\right)}{\left(x+3\right)^2\cdot\left(x-3\right)^2}\)

\(=\dfrac{\left(x-9\right)}{\left(x^2-9\right)\left(x-3\right)}\)

= ( x/(x-6)(x+6) - x-6/x(x+6) ) : 2x-6/x² + 6x + 6/6-x

=(  x²/x(x+6)(x-6) - (x -6 )(x-6)/x(x+6)(x-6)  ) : .....

= (12x -36 / x(x+6)(x-6) : 2x-6/ x² + 6x )+ 6/6-x

=6/x-6 + 6/6-x

= 6-6/ x-6

=0/x-6 

câu trước mình thiếu 6/6-x

bạn ơi kết quả bằng 6/x-6

câu trên sai , để mình giải đàng hoàng ra 

\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)

\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)

\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)

\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)

\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)

\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)

\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)

\(=\frac{x}{x-1}\)