xét \(\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}=\sqrt{\frac{b^4\left(a^2+b^2\right)^2+a^4\left(a^2+b^2\right)^2+a^4b^4}{a^4b^4\left(a^2+b^2\right)^2}}=\sqrt{\frac{a^8+b^8+2a^2b^6+a^4b^4+a^4b^4+2a^6b^2+a^4b^4}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}}\)=\(\sqrt{\frac{\left(a^4+b^4\right)^2+2a^2b^2\left(a^4+b^4\right)+a^4b^4}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}}=\sqrt{\frac{\left(a^4+b^4+a^2b^2\right)^2}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}}\)

\(A=\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}=\frac{\left(a^4+b^4\right)\left(a^2+b^2\right)^2+a^4b^4}{a^4b^4\left(a^2+b^2\right)^2}\)

\(=\frac{\left(a^4+b^4\right)\left(a^4+b^4+2a^2b^2\right)+a^4b^4}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}=\frac{\left(a^4+b^4\right)^2+2a^2b^2\left(a^4+b^4\right)+\left(a^2b^2\right)^2}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}\)

\(=\frac{\left(a^4+b^4+a^2b^2\right)^2}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}\)

\(\Rightarrow B=\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{A}\)\(=\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\frac{\left(a^2+b^2\right)^2-a^2b^2}{a^2b^2\left(a^2+b^2\right)}\)

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

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

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

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

\(\Rightarrow\sqrt{B}=\left|\frac{a^2+b^2+ab}{ab\left(a+b\right)}\right|=\frac{a^2+b^2+ab}{\left|ab\left(a+b\right)\right|}\)

Bài 1: 

Ta có:

\(\left(a-b+c\right)^3=a^3-b^3+c^3-3a^2b+3a^2c+3ab^2+3b^2c+3ac^2-3bc^2-6abc\)

\(\Rightarrow\left(\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\right)^3=\frac{1}{9}-\frac{2}{9}+\frac{4}{9}-\frac{1}{3}.\sqrt[3]{2}+\frac{1}{3}.\sqrt[3]{4}+\frac{1}{3}.\sqrt[3]{4}+\frac{2}{3}.\sqrt[3]{2}\)

\(+\frac{2}{3}.\sqrt[3]{2}-\frac{2}{3}.\sqrt[3]{4}-\frac{4}{3}=\sqrt[3]{2}-1\)

\(\Rightarrow\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\)

\(K=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\)

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

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

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

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

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

Chúc bạn học tốt !!!

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

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

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

Ai giải giúp mình bài 1 với bài 4 trước đi