Question

[Trang Nguyễn]

1) rút gọn và tìm A để A nguyên

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

giúp mk vs ạ mk cần gấp

Answer 1

[Lấp La Lấp Lánh]

\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(đk:a>0,a\ne1\right)\)

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

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+2}=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)

Để A nguyên

\(\Leftrightarrow A=\dfrac{\sqrt{a}-2}{\sqrt{a}}=1-\dfrac{2}{\sqrt{a}}\in Z\)

Do \(\sqrt{a}>0,\sqrt{a}\ne1\)

\(\Leftrightarrow\sqrt{a}\inƯ\left(2\right)=\left\{2\right\}\)

\(\Leftrightarrow a=4\)