A) Với \(x>y>0\),ta có: \(x^2+y^2< x^2+y^2+2xy=\left(x+y\right)^2\Rightarrow\frac{1}{x^2+y^2}>\frac{1}{\left(x+y\right)^2}\)

Xét: \(\frac{x^2-y^2}{x^2+y^2}>\frac{x^2-y^2}{\left(x+y\right)^2}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x-y}{x+y}\)--->ĐPCM

B) \(3^{16}+1=\left(3^{16}-1\right)+2=\left(3^8+1\right)\left(3^8-1\right)+2\)

\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)+2\)

\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)+2\)

\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)+2\)

\(>\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)--->ĐPCM

\(\Rightarrow\left(x-4\right):6=15\)

\(\Rightarrow x-4=90\)

\(\Rightarrow x=94\)

\(\left(x-4\right):6-5=10\)

\(\left(x-4\right):6=5+10\)

\(\left(x-4\right):6=15\)

\(x-4=15\cdot6\)

\(x-4=90\)

\(x=90+4\)

\(\Rightarrow x=94\)

+)  \(x^3=x^2\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

+) \((7x-11)^3=2^5.5^2+200\)

\((7x-11)^3=2^3.2^2.5^2+2^3.5^2\)

\((7x-11)^3=2^3.5^2.(2^2+1)\)

\((7x-11)^3=2^3.5^2.5\)

\((7x-11)^3=2^3.5^3\)

\((7x-11)^3=10^3\)

\(\Rightarrow7x-11=10\)

                  \(7x=21\)

                    \(x=3\)

+) \(3+2^{x-1}=24-[4^2-(2^2-1)]\)

    \(3+2^{x-1}=11\)

              \(2^{x-1}=8\)

              \(2^{x-1}=2^3\)

       \(\Rightarrow x-1=3\)

                    \(x=4\)

Ta có: \(\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)

\(=\frac{2\sqrt{3}}{3}+\frac{\sqrt{2}}{3}+\frac{2\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{12}.\sqrt{\frac{1}{12}-\frac{1}{\sqrt{6}}}\)

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{12\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)}\)

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\sqrt{5-2\sqrt{6}}\)

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

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left|\sqrt{3}-\sqrt{2}\right|\)

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left(\sqrt{3}-\sqrt{2}\right)\)(vì \(\sqrt{3}-\sqrt{2}>0\))

\(=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{3}=\sqrt{3}\)

\(\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\frac{3+6\sqrt{3}}{\sqrt{3}}-\frac{13}{\sqrt{3}+4}\)

\(=\frac{-\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\frac{4^2-\left(\sqrt{3}\right)^2}{\sqrt{3}+4}\)

\(=-\sqrt{3}+6+\sqrt{3}-\left(4-\sqrt{3}\right)\)

\(=-\sqrt{3}+6+\sqrt{3}-4+\sqrt{3}=\sqrt{3}+2\)

Ta có: \(\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)

\(=\sqrt{\frac{\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)}{\left(5-\sqrt{21}\right)^2}}+\sqrt{\frac{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}{\left(5+\sqrt{21}\right)^2}}\)

\(=\sqrt{\frac{4}{\left(5-\sqrt{21}\right)^2}}+\sqrt{\frac{4}{\left(5+\sqrt{21}\right)^2}}\)

\(=2\left(\frac{1}{5-\sqrt{21}}+\frac{1}{5+\sqrt{21}}\right)\)

\(=2.\frac{5+\sqrt{21}+5-\sqrt{21}}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}=\frac{2.10}{4}=5\)

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