Simplify $\displaystyle\frac{2^2-2}{2^2+2}\cdot\displaystyle\frac{3^2-3}{3^2+3}\cdots\displaystyle\frac{10^2-2}{10^2+10}$
Compute $\Large(\sqrt{6+4\sqrt{2}} + \sqrt{6-4\sqrt{2}}\Large)^2$
Show that $\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\cdots + \frac{1}{n\times (n+1)\times (n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$
Find the value of $x^3+x^2y+xy^2+y^3$ if $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ and $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$.
Simplify $$\sqrt{10+4\sqrt{3-2\sqrt{2}}}$$
Let $\sqrt{1+\sqrt{21+12\sqrt{3}}}=\sqrt{a}+\sqrt{b}$. Find $a+b$.
Let $a\ge 0, n\ge 0,$ and $m > 0$. Show that $\sqrt{a+m}+\sqrt{a+m+n} > \sqrt{a} + \sqrt{a+2m+n}$.
Simplify $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$
Simplify $(\sqrt[3]{3}+\sqrt[3]{2})(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})$
Simplify $\sqrt{1 + 1995\sqrt{4 + 1995 \cdot 1999}}$.
Evaluate $\sqrt{5+\sqrt{5^2+\sqrt{5^4+\sqrt{5^8+...}}}}$
Simplify $\sqrt{1-\frac{\sqrt{3}}{2}}$
Simplify $$\frac{1}{2+\frac{1}{2+\cdots}}$$
Simplify $$(\sqrt{2})^{(\sqrt{2})^{(\sqrt{2})^{\cdots}}}$$
Simplify $$2^{\sqrt{2^{\sqrt{2^{\sqrt{2}^{\cdots}}}}}}$$
Compute $$\frac{1}{\frac{1}{\frac{1}{\cdots}+1+\frac{1}{\cdots}}+1+\frac{1}{\frac{1}{\cdots}+1+\frac{1}{\cdots}}}$$
Use at least two ways to prove $$\sqrt{x\sqrt{x\sqrt{x\sqrt{\cdots}}}}=x$$
Show that $$\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=\frac{1}{\frac{1}{1+\frac{1}{1+\cdots}}}=\frac{1+\sqrt{5}}{2}$$
Show that, if both converge, $$\sqrt{a+b\sqrt{a+b\sqrt{a+\cdots}}}=b+\frac{a}{b+\frac{a}{b+\cdots}}=\frac{b+\sqrt{b^2+4a}}{2}$$
Compute $$\sqrt{\frac{2}{2^2}+\sqrt{\frac{2}{2^4}+\sqrt{\frac{2}{2^8}+\cdots}}}$$
Compute $$\sqrt{\frac{2}{2^1}+\sqrt{\frac{2}{2^2}+\sqrt{\frac{2}{2^4}+\cdots}}}$$
Compute $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
Compute $$\sqrt{6+2\sqrt{7+3\sqrt{8+\cdots}}}$$
Without using a calculator, explain that $$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}}\approx 1$$
Show that $$\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}=\frac{1+\sqrt{1+4n}}{2}$$ and $$\sqrt{n-\sqrt{n-\sqrt{n-\cdots}}}=\frac{-1+\sqrt{1+4n}}{2}$$