a, Q=\(\left(\frac{1}{\sqrt{a}+1}-\frac{1}{a+\sqrt{a}}\right):\frac{\sqrt{a}-1}{a+2\sqrt{a}+1}\) (đk: \(a>0,a\ne1\))
= \(\left(\frac{1}{\sqrt{a}+1}-\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\frac{a+2\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}=\frac{\sqrt{a}+1}{\sqrt{a}}\)
=\(1+\frac{1}{\sqrt{a}}\)
b, Để Q=0,4 <=> \(1+\frac{1}{\sqrt{a}}=0.4\)
<=> \(\frac{1}{\sqrt{a}}=-\frac{3}{5}\)(vô nghiệm)
Vậy \(a\in\left\{\varnothing\right\}\) <=> Q=0,4
c, Có Q= \(1+\frac{1}{\sqrt{a}}\)
vì \(\frac{1}{\sqrt{a}}>0\) vs mọi a thỏa mãn đk => 1+\(\frac{1}{\sqrt{a}}>1+0=1\)
<=> Q>1
Vậy Q>1
d, Có a=\(\sqrt{9-4\sqrt{5}}=\sqrt{4-4\sqrt{5}+5}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)
Thay a=\(\sqrt{5}+2\) vào Q đã rút gọn có:
Q= 1+ \(\frac{1}{\sqrt{\sqrt{5}+2}}=1+\frac{\sqrt{\sqrt{5}-2}}{\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}-2\right)}}=1+\sqrt{\sqrt{5}-2}\)
Vậy Q=\(1+\sqrt{\sqrt{5}-2}\)
e, Có Q= \(1+\frac{1}{\sqrt{a}}\)
Để Q\(\in Z\) <=> \(\frac{1}{\sqrt{a}}\in Z\) => \(\frac{1}{\sqrt{a}}\in N^+\)
Với mọi a>0 và a\(\ne1\) có: \(\left[{}\begin{matrix}\sqrt{a}\in N^+\\\sqrt{a}\notin N^+\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\frac{1}{\sqrt{a}}\in N^+\left(tm\right)\\\frac{1}{\sqrt{a}}\notin N^+\left(ktm\right)\end{matrix}\right.\)
=> \(\sqrt{a}\in\) ước TN của 1
=> \(\sqrt{a}=1\) <=> a=1
Vậy a=1 thì Q \(\in Z\)
