Question

M = \(\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\cdot\frac{\sqrt{a}+1}{\sqrt{a}}\). Rút gọn và tìm a thuộc Z để M là số nguyên giúp mình vs mn ơi

Answer 1

\(M=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}-2}{\sqrt{a}\left(\sqrt{a}-1\right)}\) ĐKXĐ :a>0, a\(\ne\) 1

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

\(=\frac{4}{a-1}\)

Để M là số nguyên\(\Leftrightarrow a-1\inƯ_{\left(4\right)}\)

\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=3\\a=2\end{matrix}\right.\)