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

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

ĐK: \(ab\ge0;b\ne0\)

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

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

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

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

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

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

a) \(A=\left[\left(\frac{1}{5}\right)^2\right]^{\frac{-3}{2}}-\left[2^{-3}\right]^{\frac{-2}{3}}=5^3-2^2=121\)

b) \(B=6^2+\left[\left(\frac{1}{5}\right)^{\frac{3}{4}}\right]^{-4}=6^2+5^3=161\)

c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}=\frac{a^{\sqrt{5}+3}.a^{5-\sqrt{5}}}{a^{\left(2\sqrt{2}\right)^2-1^2}}\)

                              \(=\frac{a^{\sqrt{5}+3+5-\sqrt{5}}}{a^{8-1}}=\frac{a^8}{a^7}=a\)

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

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

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

Ta có:\(7\left(\frac{1}{a^2}+...\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+2015\)

Mà \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le2015\)=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{6045}\)

\(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+...\)

Mà \(\left(2+1\right)\left(2a^2+b^2\right)\ge\left(2a+b\right)^2\)(bất dẳng thức buniacoxki)

=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)

Lại có \(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)

=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\le\frac{\sqrt{6045}}{3}\)

Vậy \(MaxP=\frac{\sqrt{6045}}{3}\)khi \(a=b=c=\frac{\sqrt{6045}}{2015}\)