Tính: \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x-4}{8}\)
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30 tháng 6 2017
a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
=\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\frac{x-1+x\left(x-1\right)+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)
\(=\frac{4x}{\left(x+1\right)^2}\)=VP
b.VT\(=\frac{2+x}{2-x}.\frac{\left(2-x\right)^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{4-2x+x^2}{2-x}\right)\)
=\(\frac{4-x^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{4-x^2}\right)=\frac{4-x^2}{4x^2}.\frac{2\left(2+x\right)-4}{4-x^2}\)
=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP
c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\left[\frac{3\left(x+y\right)+3x}{\left(x+y\right)\left(x-y\right)}.\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\frac{3\left(2x+y\right)\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}.\frac{x-y}{3}\)
\(=x+y=\)VP
Vậy các đẳng thức được chứng minh
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PA
21 tháng 10 2016
\(\frac{2+x}{2-x}\div\frac{4x^2}{4-4x+x^2}\times\left(\frac{2}{2-x}-\frac{8}{8+x^3}\times\frac{4-2x+x^2}{2-x}\right)\)
\(=\frac{2+x}{2-x}\times\frac{4-4x+x^2}{4x^2}\times\left(\frac{2}{2-x}-\frac{8}{\left(2+x\right)\left(4-2x+x^2\right)}\times\frac{4-2x+x^2}{2-x}\right)\)
\(=\frac{2+x}{2-x}\times\frac{\left(2-x\right)^2}{4x^2}\times\left(\frac{2\left(2+x\right)}{\left(2+x\right)\left(2+x\right)}-\frac{8}{\left(2+x\right)\left(2-x\right)}\right)\)
\(=\frac{\left(2+x\right)\left(2-x\right)}{4x^2}\times\frac{4+2x-8}{\left(2+x\right)\left(2-x\right)}\)
\(=\frac{2\left(2+x-4\right)}{4x^2}\)
\(=\frac{x-2}{2x^2}\)
26 tháng 7 2017
Ta có \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(=\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{2x}{\left(x+2\right)\left(x-2\right)}\)
\(\frac{-4.2x}{\left(x+2\right)^2\left(x-2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{2x}=\frac{-4}{\left(x+2\right)\left(x-2\right)}\)