Tính được 

\(M=\frac{6\sqrt{a}}{\left(\sqrt{a}+1\right)^2}\)

Với mọi a>0; \(a\ne1,\)ta có: \(\frac{6\sqrt{a}}{\left(\sqrt{a}+1\right)^2}>0\Leftrightarrow M>0\left(1\right)\)

Lại có:

\(a-\sqrt{a}+1>0\forall a>0\)

\(\Leftrightarrow2a+4\sqrt{a}+2>6\sqrt{a}\)\(\Rightarrow2>\frac{6\sqrt{a}}{\left(\sqrt{a}+1\right)^2}\Leftrightarrow M< 2\)(2)

Từ (1) và (2) => M đạt giá trị nguyên khi M=1

Bạn tự tìm a nha... 

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

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

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

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

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

\(N=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)

\(=\left(\frac{a+1+2\sqrt{a}-a-1+2\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)

\(=\left(\frac{4\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}=4\sqrt{a}\left(\frac{1}{a-1}+1\right)\cdot\frac{a-1}{\sqrt{a}}=4\cdot\left(a-1\right)\left(\frac{1}{a-1}+1\right)\)

\(=4\cdot\left(a-1\right)\)

vừa tham khảo cách làm vừa check lại hộ tớ với nhé :33 

\(M=(\frac{2+\sqrt{a}}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}).(\frac{a\sqrt{a}+a-\sqrt{a}-1}{\sqrt{a}})\)

\(=[\frac{\sqrt{a}+2}{(\sqrt{a}+1)^2}-\frac{\sqrt{a}-2}{(\sqrt{a}+1)(\sqrt{a}-1)}].\frac{(a\sqrt{a}-\sqrt{a})+(\sqrt{a}-1)}{\sqrt{a}}\)

\(=[\frac{(\sqrt{a}-2).(\sqrt{a}-1)}{(\sqrt{a}+1)^2.(\sqrt{a}-1)}-\frac{(\sqrt{a}-2).(\sqrt{a}+1)}{(\sqrt{a}+1)^2.(\sqrt{a}-1)}].\frac{\sqrt{a}(a-1)+(a-1)}{\sqrt{a}}\)

\(=[\frac{a+\sqrt{a}-2}{(\sqrt{a}+1)(a-1)}-\frac{a-\sqrt{a}-2}{(\sqrt{a}+1)(a-1)}].\frac{(a-1).(\sqrt{a}+1)}{\sqrt{a}}\)

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

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

\(=2\)

Vậy \(M=2\)

\(N=(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}).(\sqrt{a}-\frac{1}{\sqrt{a}})\)

\(=[\frac{(\sqrt{a}+1).(\sqrt{a}+1)}{\left(\sqrt{a}-1\right).(\sqrt{a}+1)}-\frac{(\sqrt{a}-1).(\sqrt{a}-1)}{(\sqrt{a}-1).(\sqrt{a}+1)}+\frac{4\sqrt{a}(a-1)}{(\sqrt{a}-1).(\sqrt{a}+1)}].\frac{a-1}{\sqrt{a}}\)

\(=\frac{(\sqrt{a}+1)^2-(\sqrt{a}-1)^2+(4a\sqrt{a}-4\sqrt{a})}{(\sqrt{a}-1).(\sqrt{a}+1)}.\frac{a-1}{\sqrt{a}}\)

\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}}\)

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

Vậy \(N=4a\)