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\(\left(\frac{x}{x^2-36}+\frac{6-x}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
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
\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)
\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)
\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)
\(=\frac{x}{x-1}\)
A = \(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
= \(\left[\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\right]:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\left[\frac{x^2}{x\left(x-6\right)\left(x+6\right)}-\frac{\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right]:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\frac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\frac{\left(x-x+6\right)\left(x+x-6\right)}{x\left(x-6\right)\left(x+6\right)}:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
=
= \(\frac{x\left(2x-6\right)}{x\left(x-6\right)\left(x+6\right)}:\frac{2x-6}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\frac{2x-6}{\left(x-6\right)\left(x+6\right)}.\frac{x\left(x+6\right)}{2x-6}\) \(-\frac{x}{x-6}\)
= \(\frac{x}{x-6}-\frac{x}{x-6}\)
= 0