Đáp án:
Giải thích các bước giải:
10,
Áp dụng:
\[\begin{array}{l}
{x^2} + {y^2} + {z^2} = {\left( {x + y + z} \right)^2} - 2\left( {xy + yz + zx} \right)\\
\Rightarrow \left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} \right) = {\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)^2} - 2\left( {\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}}} \right)\\
= {\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)^2} - 2.\frac{{a + b + c}}{{abc}}\\
= {\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)^2}
\end{array}\]
21
Ta có
\[\begin{array}{l}
{\left( {1 + \frac{1}{a} - \frac{1}{{a + 1}}} \right)^2} = 1 + \frac{1}{{{a^2}}} + \frac{1}{{{{\left( {a + 1} \right)}^2}}} + \frac{2}{a} - \frac{2}{{a\left( {a + 1} \right)}} - \frac{2}{{a + 1}}\\
= 1 + \frac{1}{{{a^2}}} + \frac{1}{{{{\left( {a + 1} \right)}^2}}} + \frac{2}{a} - \left( {\frac{2}{a} - \frac{2}{{a + 1}}} \right) - \frac{2}{{a + 1}}\\
= 1 + \frac{1}{{{a^2}}} + \frac{1}{{{{\left( {a + 1} \right)}^2}}}
\end{array}\]
28
Áp dung:
\[\begin{array}{l}
\left( {\sqrt {n - 1} + \sqrt n } \right)\left( {\sqrt n - \sqrt {n - 1} } \right)\\
= \left( {n - \left( {n - 1} \right)} \right)\\
= 1\\
\Rightarrow \frac{1}{{\sqrt {n - 1} + \sqrt n }} = \sqrt n - \sqrt {n - 1}
\end{array}\]
