Question

Cho A =\(\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\) với a>0 và a\(\ne\)1

a, Rút gọn A

b,So sánh giá trị của A với 1

Answer 1

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

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

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

\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)

vậy ....

b)\(đk:a>0;a\ne1\)

xét hiệu A-1 có \(\frac{\sqrt{a}-1}{\sqrt{a}}-1=\frac{\sqrt{a}-1-\sqrt{a}}{\sqrt{a}}=-\frac{1}{\sqrt{a}}\)

có \(a>0\forall atmđk=>\sqrt{a}>0\forall atmđk=>-\frac{1}{\sqrt{a}}< -1\forall atmđk\)

=> A-1<0=> A<1

Answer 2

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

b) Theo câu a ta có: \(A=1-\frac{1}{\sqrt{a}}< 1\)( vì \(\frac{1}{\sqrt{a}}>0\))