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d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{8}{x^2-1}\right):\left(\frac{1}{x-1}-\frac{7x+3}{1-x^2}\right)\)
\(A=\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x+1\right)\left(x-1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}\right]:\left[\frac{x+1}{\left(x+1\right)\left(x-1\right)}-\frac{3-7x}{\left(x+1\right)\left(x-1\right)}\right]\)
\(A=\left[\frac{x^2+2x+1-x^2+2x-1+8}{\left(x+1\right)\left(x-1\right)}\right]:\frac{x+1-3+7x}{\left(x+1\right)\left(x-1\right)}\)
\(A=\frac{4x+8}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{8x-2}\)
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\(E=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right).\left(\frac{1-x^2}{2}\right)^2\)
\(E=\left(\frac{x-2}{\left(x-1\right)\left(x+1\right)}-\frac{x-2}{\left(x+1\right)^2}\right).\left(\frac{\left(1-x\right)\left(1+x\right)}{2}\right)^2\)
\(E=\left(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)^2}-\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)^2}\right).\frac{\left(1-x\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{\left(x-2\right)\left(x+1-x+1\right)}{\left(x-1\right)\left(x+1\right)^2}.\frac{\left(x-1\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{2\left(x-2\right)\left(x-1\right)}{4}\)
\(E=\frac{\left(x-2\right)\left(x-1\right)}{2}\)
a) \(E=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right).\left(\frac{1-x^2}{2}\right)^2\)
\(=\left(\frac{x-2}{\left(x-1\right)\left(x+1\right)}-\frac{x+2}{\left(x+1\right)^2}\right).\frac{\left(x^2-1\right)^2}{4}\)
\(=\left(\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\right).\frac{\left(x^2-1\right)^2}{4}\)
\(=\left(\frac{x^2-3x+2-x^2-3x-2}{\left(x-1\right)^2\left(x+1\right)}\right).\frac{\left(x^2-1\right)^2}{4}\)
\(=\frac{-6x.\left(x^2-1\right)^2}{\left(x-1\right)^2\left(x+1\right).4}=\frac{-3x\left(x^2-1\right)^2}{\left(x^2-1\right)\left(x-1\right).4}=\frac{-3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right).4}\)\(=\frac{-3x\left(x+1\right)}{4}\)
b) Muốn \(\frac{E-4}{5}=x\) thì \(\frac{\frac{-3x\left(x+1\right)}{4}-4}{5}=x\)
\(\Rightarrow\frac{\frac{-3x^2\left(x+1\right)}{4}-\frac{16}{4}}{5}=x\)
\(\Rightarrow\frac{-3x^3-3x^2-16}{4}=5x\)
\(\Rightarrow-3x^3-3x^2-16=20x\)
\(\Rightarrow-3x^3-3x^2-16=20x\).....................................................................