\(M=\frac{\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)^2}{\sqrt[3]{ab}}:\left(2+\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\right)=\frac{\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)^2}{\sqrt[3]{ab}}:\frac{2\sqrt[3]{ab}+\left(\sqrt[3]{a}\right)^2+\left(\sqrt[3]{a}\right)^2}{\sqrt[3]{ab}}\)

    \(=\frac{\left(\sqrt[3]{a}+\sqrt[3]{b}\right)^2}{\sqrt[3]{ab}}-\frac{\sqrt[3]{ab}}{\left(\sqrt[3]{a}+\sqrt[3]{b}\right)^2}=1\)

\(=\left(\sqrt{a}-\sqrt{b}\right)^2:\left(\sqrt{b}-\frac{b}{\sqrt{a}}\right)^2=\left(\sqrt{a}-\sqrt{b}\right)^2:\left[\frac{\sqrt{b}}{\sqrt{a}}\left(\sqrt{a}-\sqrt{b}\right)\right]^2\)

\(=\left(\sqrt{a}-\sqrt{b}\right)^2:\left[\frac{\sqrt{b}}{\sqrt{a}}\left(\sqrt{a}-\sqrt{b}\right)\right]^2\)

 \(=\left(\sqrt{a}-\sqrt{b}\right)^2.\frac{a}{b\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{a}{b}\)

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

                                                        \(=\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{b}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{1}{b}\)

ĐK: \(ab\ge0;b\ne0\)

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

\(=\left(\sqrt{\frac{a}{b}}-1\right)^2:\left(\sqrt{a}-\sqrt{b}\right)^2=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{b}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{1}{b}\)

a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)

b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)

c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)

\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)

Lời giải:

a) ĐK: $a\neq -b\neq 0$

\(A=\left(\frac{a^2+b^2}{a^2b^2}+\frac{2}{a+b}.\frac{a+b}{ab}\right).\frac{ab}{(a+b)^2}\)

\(=\left(\frac{a^2+b^2}{a^2b^2}+\frac{2ab}{a^2b^2}\right).\frac{ab}{(a+b)^2}=\frac{(a+b)^2}{a^2b^2}.\frac{ab}{(a+b)^2}=\frac{1}{ab}\)

b)

\(B=\left[\frac{(2x+y)^2}{(2x-y)^2(2x+y)^2}+\frac{(2x-y)^2}{(2x-y)^2(2x+y)^2}+\frac{2}{(2x-y)(2x+y)}\right].\frac{(2x+y)^2}{16x}\)

\(=\left[\frac{8x^2+2y^2}{(2x-y)^2(2x+y)^2}+\frac{2(2x-y)(2x+y)}{(2x-y)^2(2x+y)^2}\right].\frac{(2x+y)^2}{16x}\)

\(=\frac{8x^2+2y^2+2(4x^2-y^2)}{(2x-y)^2(2x+y)^2}.\frac{(2x+y)^2}{16x}\)

\(=\frac{16x^2}{(2x-y)^2(2x+y)^2}.\frac{(2x+y)^2}{16x}=\frac{x}{(2x-y)^2}\)