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\(A=\left(\frac{2x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{5-x^2}{x+2}\right)\) ĐKXĐ : \(x\ne\pm2\)
\(A=\left(\frac{2x}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4}{x+2}+\frac{5-x^2}{x+2}\right)\)
\(A=\left(\frac{2x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+5-x^2}{x+2}\right)\)
\(A=\frac{x-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1}\)
\(A=\frac{x-6}{x-2}\)
\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\) ĐK đề bài
\(=\frac{x-5+2\left(x+5\right)-2x-10}{\left(x+5\right)\left(x-5\right)}=\frac{-\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=-\frac{1}{x-5}\)
b/ có A=-3 => \(-\frac{1}{x-5}=-3 \Rightarrow x-5=\frac{1}{3}\Rightarrow x=\frac{16}{3}\)
có \(9x^2-42x+49=\left(3x-7\right)^2=\left(\frac{3.16}{3}-7\right)^2=81\)
\(E=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right).\left(\frac{1-x^2}{2}\right)^2\)
\(E=\left(\frac{x-2}{\left(x-1\right)\left(x+1\right)}-\frac{x-2}{\left(x+1\right)^2}\right).\left(\frac{\left(1-x\right)\left(1+x\right)}{2}\right)^2\)
\(E=\left(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)^2}-\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)^2}\right).\frac{\left(1-x\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{\left(x-2\right)\left(x+1-x+1\right)}{\left(x-1\right)\left(x+1\right)^2}.\frac{\left(x-1\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{2\left(x-2\right)\left(x-1\right)}{4}\)
\(E=\frac{\left(x-2\right)\left(x-1\right)}{2}\)
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 \(\frac{\frac{-3x\left(x+1\right)}{4}-4}{5}=x\)
\(\Rightarrow\frac{\frac{-3x^2\left(x+1\right)}{4}-\frac{16}{4}}{5}=x\)
\(\Rightarrow\frac{-3x^3-3x^2-16}{4}=5x\)
\(\Rightarrow-3x^3-3x^2-16=20x\)
\(\Rightarrow-3x^3-3x^2-16=20x\).....................................................................
\(a)\dfrac{3}{{x + 5}} + \dfrac{2}{{x - 5}} - \dfrac{{2x + 10}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} = \dfrac{3}{{x + 5}} + \dfrac{2}{{x - 5}} - \dfrac{{2x + 10}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}\\ = \dfrac{3}{{x + 5}} + \dfrac{2}{{x - 5}} - \dfrac{{2\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}\\ = \dfrac{3}{{x + 5}} + \dfrac{2}{{x - 5}} - \dfrac{2}{{x - 5}}\\ = \dfrac{3}{{x + 5}}\\ \)