Cho a + b khác 0; a; b khác 0. Chứng minh rằng: căn bậc hia của 1/a^2 + 1/b^2 + 1/( a + b)^2  = | 1/a + 1/b - 1/a + b|

Câu hỏi:

Trả lời:

Lời giải

Ta có $\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{{\left( {a + b} \right)}^2}}}}$

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

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

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

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

$= \left| {\frac{1}{a} + \frac{1}{b} - \frac{1}{{a + b}}} \right|$.

Vậy $\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{{\left( {a + b} \right)}^2}}}} = \left| {\frac{1}{a} + \frac{1}{b} - \frac{1}{{a + b}}} \right|$ (đpcm).