Bài 1: 

a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)

b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)

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

đkxđ: m≠0, n ≠ 0; mn > 0; m ≠ \(\sqrt{mn}\)

\(\dfrac{m+n}{\sqrt{m}+\sqrt{n}}:\left(\dfrac{m+n}{\sqrt{mn}}+\dfrac{n}{m-\sqrt{mn}}-\dfrac{m}{n+\sqrt{mn}}\right)\)

\(=\dfrac{m+n}{\sqrt{m}+\sqrt{n}}:\left(\dfrac{m+n}{\sqrt{mn}}+\dfrac{n}{\sqrt{m}\left(\sqrt{m}-\sqrt{n}\right)}-\dfrac{m}{\sqrt{n}\left(\sqrt{m}+\sqrt{n}\right)}\right)\)

\(=\dfrac{m+n}{\sqrt{m}+\sqrt{n}}:\left[\dfrac{\left(m+n\right)\left(m-n\right)}{\sqrt{mn}\left(m-n\right)}+\dfrac{n\sqrt{n}\left(\sqrt{m}+\sqrt{n}\right)}{\sqrt{mn}\left(m-n\right)}-\dfrac{m\sqrt{m}\left(\sqrt{m}-\sqrt{n}\right)}{\sqrt{mn}\left(m-n\right)}\right]\)

\(=\dfrac{m+n}{\sqrt{m}+\sqrt{n}}:\dfrac{m^2-n^2+n\sqrt{mn}+n^2-m^2+m\sqrt{mn}}{\sqrt{mn}\left(m-n\right)}\)

\(=\dfrac{m+n}{\sqrt{m}+\sqrt{n}}:\dfrac{n\sqrt{mn}+m\sqrt{mn}}{\sqrt{mn}\left(m-n\right)}\)

\(=\dfrac{m+n}{\sqrt{m}+\sqrt{n}}\cdot\dfrac{\sqrt{mn}\left(\sqrt{m}-\sqrt{n}\right)\left(\sqrt{m}+\sqrt{n}\right)}{\sqrt{mn}\left(m+n\right)}\)

\(=\sqrt{m}-\sqrt{n}\)

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

\(=ab+2b-\sqrt{ab\cdot\dfrac{a\sqrt{a}+\sqrt{b}}{b\sqrt{a}}}\)

\(=ab+2b-\sqrt{\sqrt{a}\cdot\left(a\sqrt{a}+\sqrt{b}\right)}\)

b: \(=\left(\sqrt{\dfrac{a^2m^2\cdot n}{b^2\cdot m}}-\sqrt{mn\cdot\dfrac{a^2b^2}{n^2}}+\sqrt{\dfrac{a^4}{b^4}\cdot\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)

\(=\left(\dfrac{a\sqrt{mn}}{b}-\sqrt{a^2b^2\cdot\dfrac{m}{n}}+\dfrac{a^2}{b^2}\cdot\sqrt{\dfrac{m}{n}}\right)\cdot\sqrt{\dfrac{n}{m}}\cdot a^2b^2\)

\(=\left(\dfrac{an}{b}-ab+\dfrac{a^2}{b^2}\right)\cdot a^2b^2\)

\(=a^3nb-a^3b^3+a^4\)

Bài 2: 

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

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

b: Để P>=-2 thì P+2>=0

\(\Leftrightarrow-2\sqrt{a}+2>=0\)

=>0<=a<1

a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)

\(M=\sqrt{3}-\sqrt{3}+1\)

\(M=1\)

b) Ta có:

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

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

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

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

\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Theo đề ta có: \(M=2N\)

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

\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)

\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)

\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)

\(\Leftrightarrow\sqrt{a}=2\)

\(\Leftrightarrow a=4\left(tm\right)\)