Đơn giản biểu thức :\(A=\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}+\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{b}{\left(a+b\right)^5}+\left(\frac{1}{a}+\frac{1}{b}\right)\)
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11 tháng 5 2016
\(=\left[\frac{\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\left(a+a^{\frac{1}{2}}b^{\frac{1}{2}}+b\right)}{a^{\frac{1}{2}}-b^{\frac{1}{2}}}+a^{\frac{1}{2}}b^{\frac{1}{2}}\right]\left[\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}\right]^2\)
\(=\frac{a+2a^{\frac{1}{2}}b^{\frac{1}{2}}+b}{\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)^2}=\frac{\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)^2}{\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)^2}=1\)
3 tháng 8 2020
Ta có : \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{1}{a}+\frac{1}{b}\right)\)
=> \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{a^3+b^3}{a^3b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{a^2+b^2}{a^2b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{a+b}{ab}\right)\)
=> \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{a^3+b^3}{1}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{a^2+b^2}{1}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{a+b}{1}\right)\)
=> \(P=\frac{a^3+b^3}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)
=> \(P=\frac{\left(a+b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2+2a\right)-6a}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)
=> \(P=\frac{\left(a+b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)^3}+\frac{3\left(a+b\right)^2}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}-\frac{6}{\left(a+b\right)^4}\)
=> \(P=\frac{a^2+ab+b^2}{\left(a+b\right)^2}+\frac{3}{\left(a+b\right)^2}+\frac{6}{\left(a+b\right)^4}-\frac{6}{\left(a+b\right)^4}\)
=> \(P=\frac{a^2+ab+b^2}{\left(a+b\right)^2}+\frac{3}{\left(a+b\right)^2}=\frac{2a^2+4ab+2b^2}{\left(a+b\right)^2}-\frac{a^2+b^2}{\left(a+b\right)^2}\)
=> \(P=2-\frac{a^2+b^2}{\left(a+b\right)^2}=1+\frac{-2}{\left(a+b\right)^2}\)
11 tháng 5 2016
\(M=\frac{\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)^2}{\sqrt[3]{ab}}:\left(2+\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\right)=\frac{\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)^2}{\sqrt[3]{ab}}:\frac{2\sqrt[3]{ab}+\left(\sqrt[3]{a}\right)^2+\left(\sqrt[3]{a}\right)^2}{\sqrt[3]{ab}}\)
\(=\frac{\left(\sqrt[3]{a}+\sqrt[3]{b}\right)^2}{\sqrt[3]{ab}}-\frac{\sqrt[3]{ab}}{\left(\sqrt[3]{a}+\sqrt[3]{b}\right)^2}=1\)
TM
28 tháng 9 2017
\(A=\frac{1}{\left(a+b\right)^3}.\frac{a^3+b^3}{\left(ab\right)^3}+\frac{3}{\left(a+b\right)^4}.\frac{a^2+b^2}{\left(ab\right)^2}+\frac{6}{\left(a+b\right)^5}.\frac{a+b}{ab}\)
\(=\frac{1}{\left(a+b\right)^3}.\frac{a^3+b^3}{1^3}+\frac{3}{\left(a+b\right)^4}.\frac{a^2+b^2}{1^2}+\frac{6}{\left(a+b\right)^5}.\frac{a+b}{1}\)
\(=\frac{a^2-ab+b^2}{\left(a+b\right)^2}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6}{\left(a+b\right)^4}\)\(=\frac{\left(a^3+b^3\right)\left(a+b\right)+3a^2+3b^2+6}{\left(a+b\right)^4}\)
\(=\frac{a^4+a^3b+ab^3+b^4+3a^2+3b^2+6}{a^4+4a^3b+6a^2b^2+4ab^3+b^4}\)\(=\frac{a^4+a^2.1+1.b^2+b^4+3a^2+3b^2+6}{a^4+4a^2.1+6.1^2+4b^2.1+b^4}\)
\(=\frac{a^4+4a^2+4b^2+b^4+6}{a^4+4a^2+6+4b^2+b^4}=1\)