đkxđ:a>0; \(a\ne1\)
,\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2.\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1-\sqrt{a}-1\right)\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\)
\(=\frac{\left(a-1\right)^2}{4a}.\frac{-2.2\sqrt{a}}{a-1}=\frac{-\left(a-1\right)}{\sqrt{a}}=\frac{1}{\sqrt{a}}-\sqrt{a}=\frac{1-a}{\sqrt{a}}\)
d, để A=-2
\(\frac{1-a}{\sqrt{a}}=-2< =>1-a=-2\sqrt{a}< =>a-2\sqrt{a}-1=0\)
đặt t=\(\sqrt{a}\left(t>0\right)\)
\(=>t^2-2t-1=0\)
ta có \(\Delta=\left(-2\right)^2-4.1.\left(-1\right)=8>0\)
vậy phương trình có 2 nghiệm phân biệt
\(t1=\frac{2+\sqrt{8}}{2.1}=1+\sqrt{2}\)
\(t2=\frac{2-\sqrt{8}}{2.1}=1-\sqrt{2}\)(loại)
với t1=\(1+\sqrt{2}\) thì \(\sqrt{a}=1+\sqrt{2}< =>a=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\)
