\(A=\frac{\frac{1-\sqrt{5}}{1+\sqrt{5}}-\frac{1+\sqrt{5}}{1-\sqrt{5}}}{\sqrt{5}}=\frac{\frac{6-2\sqrt{5}-6-2\sqrt{5}}{1-5}}{\sqrt{5}}\)

\(=\frac{\frac{4\sqrt{5}}{4}}{\sqrt{5}}=\frac{\sqrt{5}}{\sqrt{5}}=1\)

Ta có: \(2a+b^2=2a\left(a+b+c\right)+b^2=b^2+2a^2+2ab+2ac\)

\(\ge4ab+2ac+a^2\)

\(\Rightarrow\frac{a}{2a+b^2}\le\frac{a}{4ab+2ac+a^2}=\frac{1}{4b+2c+a}\)

\(\le\frac{1}{49}.\frac{49}{4b+2c+a}=\frac{1}{49}.\frac{\left(4+2+1\right)^2}{4b+2c+a}\)

\(\le\frac{1}{49}\left(\frac{16}{4b}+\frac{4}{2c}+\frac{1}{a}\right)=\frac{1}{49}\left(\frac{4}{b}+\frac{2}{c}+\frac{1}{a}\right)\)

CMTT: \(\frac{b}{2b+c^2}\le\frac{1}{49}\left(\frac{4}{c}+\frac{2}{a}+\frac{1}{b}\right);\frac{c}{2c+a^2}\le\frac{1}{49}\left(\frac{4}{a}+\frac{2}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2}\le\frac{1}{7}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)( đpcm )

Ta có \(\sqrt{x}\ge0\forall x\ne0;x\ne1\Rightarrow-5\sqrt{x}\le0\Rightarrow2-5\sqrt{x}\le2\)

\(\Rightarrow A\le\frac{2}{3}\forall x\ge0;x\ne1\)

Vậy GTLN của A  là 2/3 <=> x=0

3. a) \(M=3x-\sqrt[3]{27x^3+27x^2+9x+1}\)

\(=3x-\sqrt[3]{\left(3x\right)^3+3.\left(3x\right)^2.1+3.\left(3x\right).1^2+1}\)

\(=3x-\sqrt[3]{\left(3x+1\right)^3}\)

\(=3x-\left(3x+1\right)\)

\(=-1\)

b) \(N=\sqrt[3]{8x^3+12x^2+6x+1}-\sqrt[3]{x^3}\)

\(=\sqrt[3]{\left(2x\right)^3+3.\left(2x\right)^2.1+3.\left(2x\right).1^2+1^3}-x\)

\(=\sqrt[3]{\left(2x+1\right)^3}-x\)

\(=2x+1-x\)

\(=x+1\)

4. a) \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)

\(=\sqrt[3]{\left(\sqrt{3}-1\right)^2\left(\sqrt{3}-1\right)}\)

\(=\sqrt[3]{\left(\sqrt{3}-1\right)^3}\)

\(=\sqrt{3}-1\)

b) \(\sqrt{3+\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}}\)

\(=\sqrt{3+\sqrt{3}+\sqrt[3]{3\sqrt{3}+3.\left(\sqrt{3}\right)^2.1+3.\sqrt{3}.1^2+1}}\)

\(=\sqrt{3+\sqrt{3}+\sqrt[3]{\left(\sqrt{3}+1\right)^3}}\)

\(=\sqrt{3+\sqrt{3}+\sqrt{3}+1}\)

\(=\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left|\sqrt{3}+1\right|\)

\(=\sqrt{3}+1\)(do \(\sqrt{3};1>0\))

1. a) \(\sqrt[3]{512}=\sqrt[3]{8^3}=8\)

b) \(\sqrt[3]{\frac{-1}{125}}=\sqrt[3]{\left(-\frac{1}{5}\right)^3}=-\frac{1}{5}\)

c) \(\sqrt[3]{\frac{343a^3b^6}{-216}}=\sqrt[3]{\left(\frac{7ab^2}{-6}\right)^3}=\frac{7ab^2}{-6}=-\frac{7ab^2}{6}\)

d) \(\sqrt[3]{-64a^9b^9}=\sqrt[3]{\left(-2a^3b^3\right)^3}=-2a^3b^3\)

2. a) \(\frac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)

\(=\sqrt[3]{\frac{135}{5}}-\sqrt[3]{54.4}\)

\(=\sqrt[3]{27}-\sqrt[3]{216}\)

\(=\sqrt[3]{3^3}-\sqrt[3]{6^3}\)

\(=3-6=-3\)

b) \(\left(\sqrt[3]{25}-\sqrt[3]{10}+\sqrt[3]{4}\right)\left(\sqrt[3]{5}+\sqrt[3]{2}\right)\)

\(=\sqrt[3]{25}.\sqrt[2]{5}+\sqrt[3]{25}.\sqrt[3]{2}-\sqrt[3]{10}.\sqrt[3]{5}-\sqrt[3]{10}.\sqrt[3]{2}+\sqrt[3]{4}.\sqrt[3]{5}+\sqrt[3]{4}.\sqrt[3]{2}\)

\(=\sqrt[3]{25.5}+\sqrt[3]{25.2}-\sqrt[3]{10.5}-\sqrt[3]{10.2}+\sqrt[3]{4.5}+\sqrt[3]{4.2}\)

\(=\sqrt[3]{125}+\sqrt[3]{50}-\sqrt[3]{50}-\sqrt[3]{20}+\sqrt[3]{20}+\sqrt[3]{8}\)

\(=\sqrt[3]{5^3}+\sqrt[3]{2^3}\)

\(=5+2=7\)