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\(C=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{\left(a+1\right)^3}\cdot\dfrac{a+1}{a}\right):\dfrac{a-1}{a^3}\)

\(=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{a\left(a+1\right)^2}\right):\dfrac{a-1}{a^3}\)

\(=\dfrac{a+2\cdot\left(a^2+1\right)}{a\left(a^2+1\right)\left(a+1\right)^2}\cdot\dfrac{a^3}{a-1}\)

\(=\dfrac{2a\left(a+1\right)}{\left(a^2+1\right)\cdot\left(a+1\right)^3}\cdot\dfrac{a^2}{a-1}\)

\(=\dfrac{2a^3}{\left(a^2+1\right)\left(a+1\right)^2\cdot\left(a-1\right)}\)

\(A=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{\left(2a-1\right)^2}{2a+1}\cdot\dfrac{1}{\left(2a-1\right)\left(2a+1\right)}\right)\cdot\left(\dfrac{4a\left(a+1\right)+1}{4a^2}\right)-\dfrac{1}{2a}\)

\(=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{2a-1}{\left(2a+1\right)^2}\right)\cdot\dfrac{4a^2+4a+1}{4a^2}-\dfrac{1}{2a}\)

\(=\dfrac{-\left(2a-1\right)\left(2a+1\right)}{\left(2a+1\right)^2}\cdot\dfrac{\left(2a+1\right)^2}{4a^2}-\dfrac{1}{2a}\)

\(=\dfrac{-\left(4a^2-1\right)}{4a^2}-\dfrac{2a}{4a^2}\)

\(=\dfrac{-4a^2-2a+1}{4a^2}\)

a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)

\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)

\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)

23 tháng 8 2017

\(A=\left[1:\left(1-\frac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)

\(=\left[1:\left(\frac{1+\sqrt{a}-\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)

\(=\left(1:\frac{1}{1+\sqrt{a}}\right).\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)

\(=\left(\sqrt{a}+1\right).\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)

\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\frac{a-1}{a+1}\)

23 tháng 9 2017

Ta có: \(B=[\left(b+c\right)^2-a^2].\)\(\dfrac{\dfrac{a+b+c}{b+c}}{\dfrac{b+c-a}{b+c}}\)\(.\)\(\dfrac{2bc}{a+b+c}\)

\(=\left(b+c+a\right)\left(b+c-a\right).\dfrac{a+b+c}{b+c-a}.\dfrac{2bc}{a+b+c}\)

\(=\left(a+b+c\right).2bc=2abc+2b^2c+2bc^2\)