Question

Giải pt: \(\sqrt{a}\) + a=1

Answer 1

ĐKXĐ: \(a\ge0\) 

\(\Leftrightarrow a+\sqrt{a}-1=0\Leftrightarrow a+2\cdot\sqrt{a}\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{5}{4}=0\Leftrightarrow\left(\sqrt{a}+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\) 

\(\Rightarrow\sqrt{a}+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\)(Vì \(\sqrt{a}+\dfrac{1}{2}\ge\dfrac{1}{2}>0\))

\(\Rightarrow\sqrt{a}=\dfrac{\sqrt{5}-1}{2}\Rightarrow a=\dfrac{6-2\sqrt{5}}{4}=\dfrac{3-\sqrt{5}}{2}\left(TM\right)\)