Question

K=\(\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a+1}}+\frac{2}{a-1}\right)\)                                               

a,rút gọn K,tính K khi a=\(3+2\sqrt{2}\)

b, tìm a để K<0

Answer 1

đk: a>0; a khác 1

\(K=\left(\frac{\sqrt{a}.\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{a-1}{\sqrt{a}+1}\)

\(=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{a-1}{\sqrt{a}+1}=\frac{a-1}{\sqrt{a}}\)

\(a=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\). thay a vào rồi tính thôi nha

b) \(K=\frac{a-1}{\sqrt{a}}\). ta thấy \(a>0\Rightarrow\sqrt{a}>0\) => K<0 <=> a-1<0 <=> a<1 => 0<a<1