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 !!!

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

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

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

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

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

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

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

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

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

Ai giải giúp mình bài 1 với bài 4 trước đi

\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)

   \(=\left[\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{b}{a}}\)

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