\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}\)
\(=\sqrt{1+a^2+\left(\dfrac{a}{a+1}\right)^2+2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}}\)
(vì \(2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}=\dfrac{2a^2+2a-2a-2a^2}{a+1}=0\))
\(=\sqrt{\left(1+a-\dfrac{a}{a+1}\right)^2}\)
\(=\left|1+a-\dfrac{a}{a+1}\right|\)
- - -
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2014^2+\dfrac{2014^2}{2015^2}}+\dfrac{2014}{2015}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=\left|1+2014-\dfrac{2014}{2015}\right|+\dfrac{2014}{2015}\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2015\)
Tới đây bn làm bảng xét dấu nhé ~^^~
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