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[chit_in]

Đơn giản biểu thức A = $\dfrac{1}{\left ( a+b \right )^{3}}\left ( \dfrac{1}{a^{4}}-\dfrac{1}{b^{4}} \right )+\dfrac{2}{\left ( a+b \right )^{4}}\left ( \dfrac{1}{a^{3}}-\dfrac{1}{b^{3}} \right )+\dfrac{2}{\left ( a+b\right )^{5}}\left ( \dfrac{1}{a^{2}}-\dfrac{1}{b^{2}} \right )$

Bài viết đã được chỉnh sửa nội dung bởi chit_in: 07-01-2012 - 22:05

[dark templar]

Đơn giản biểu thức A = $\dfrac{1}{\left ( a+b \right )^{3}}\left ( \dfrac{1}{a^{4}}-\dfrac{1}{b^{4}} \right )+\dfrac{2}{\left ( a+b \right )^{4}}\left ( \dfrac{1}{a^{3}}-\dfrac{1}{b^{3}} \right )+\dfrac{2}{\left ( a+b\right )^{5}}\left ( \dfrac{1}{a^{2}}-\dfrac{1}{b^{2}} \right )$

\[\begin{array}{l}
A = \frac{1}{{{{(a + b)}^3}}}\left( {\frac{1}{a} + \frac{1}{b}} \right)\left[ {\left( {\frac{1}{a} + \frac{1}{b}} \right)\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \right) + \frac{2}{{a + b}}\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{ab}}} \right) + \frac{2}{{{{(a + b)}^2}}}\left( {\frac{1}{a} + \frac{1}{b}} \right)} \right] \\
= \frac{1}{{{{(a + b)}^3}}}\left( {\frac{1}{a} + \frac{1}{b}} \right)\left[ {\left( {\frac{1}{a} + \frac{1}{b}} \right)\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \right) + \frac{2}{{a + b}}\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{ab}}} \right) + \frac{2}{{ab(a + b)}}} \right] \\
= \frac{1}{{{{(a + b)}^3}}}\left( {\frac{1}{a} + \frac{1}{b}} \right)\left[ {\left( {\frac{1}{a} + \frac{1}{b}} \right)\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \right) + \frac{2}{{a + b}}{{\left( {\frac{1}{a} + \frac{1}{b}} \right)}^2}} \right] \\
= \frac{1}{{{{(a + b)}^3}}}\left( {\frac{1}{{{a^2}}} - \frac{1}{{{b^2}}}} \right)\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{2}{{ab}}} \right) \\
= \frac{1}{{{{(a + b)}^3}}}\left( {\frac{1}{{{a^2}}} - \frac{1}{{{b^2}}}} \right){\left( {\frac{1}{a} + \frac{1}{b}} \right)^2} \\
\end{array}\]

Bài viết đã được chỉnh sửa nội dung bởi perfectstrong: 08-01-2012 - 14:29