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

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

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

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

Câu A dưới mẫu em ghi nhầm nha các pro
Mẫu là \(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\)

Điều kiện: \(\left\{{}\begin{matrix}b\ne0,m\ne0,n\ne0\\\dfrac{n}{m}\ge0\\mn\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b\ne0\\m\ne0\\n\ne0\\m,n\end{matrix}\right.\) ( bổ sung chổ m,n nha chỗ đó là m,n cùng dấu )

Khi đó \(A=\left(\dfrac{am}{b}\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}\right).a^2.b^2\sqrt{\dfrac{n}{m}}\)

\(=\dfrac{am}{b}\sqrt{\dfrac{n}{m}}.a^2.b^2.\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}.a^2.b^2.\sqrt{\dfrac{n}{m}}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}.a^2.b^2.\sqrt{\dfrac{n}{m}}\)

\(=a^3bm\sqrt{\dfrac{n^2}{m^2}}-\dfrac{a^3b^3}{n}.\sqrt{\dfrac{mn^2}{m}}+a^4.\sqrt{\dfrac{mn}{nm}}\)

\(=a^3bm.\left|\dfrac{n}{m}\right|-\dfrac{a^3b^3}{n}.\sqrt{n^2}+a^4\)

\(=a^3bm.\dfrac{n}{m}-\dfrac{a^3b^3}{n}.\left|n\right|+a^4\) ( vì m,n cùng dấu )

\(=a^3bn-\dfrac{a^3b^3}{n}.\left|n\right|+a^4\)

Nếu n > 0 thì \(A=a^3bn-\dfrac{a^3b^3}{n}.n+a^4=a^3bn-a^3b^3+a^4\)

Nếu n < 0 thì \(A=a^3bn-\dfrac{a^3b^3}{n}.\left(-n\right)+a^4=a^3bn+a^3b^3+a^4\)

Vậy \(A=a^3bn-a^3b^3+a^4\) với n > 0; \(A=a^3bn+a^3b^3+a^4\) với n < 0

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Èo. Câu này nhân vô đi bác. Nhân vô rồi rút gọn

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