Question

Cho a,b và a+b khác 0.cmr:

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

Answer 1

Lời giải:

Ta thấy:

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{(a+b)^2}=\frac{a^2+b^2}{a^2b^2}+\frac{1}{(a+b)^2}=\frac{(a+b)^2-2ab}{a^2b^2}+\frac{1}{(a+b)^2}\)

\(=\frac{(a+b)^2}{a^2b^2}-\frac{2}{ab}+\frac{1}{(a+b)^2}\)

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

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

\(\Rightarrow \sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{(a+b)^2}}=|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}|\)