Practice With Solutions

back to index  |  new

Let $x, y, z$ be three distinct positive real numbers such that $x+\sqrt{y+\sqrt{z}}=z+\sqrt{y+\sqrt{x}}$. Show that $40xz<1$

(Rearrangement Theorem) Let $a_1, a_2, \cdots, a_n$ and $b_1, b_2, \cdots, b_n$ be sequences of positive real numbers, and let $c_1, c_2, \cdots, c_n$ be a permutation of $b_1, b_2, \cdots, b_n$. The sum $S=a_1b_1+a_2b_2+\cdots+a_nb_n$ is maximal if the two sequences $a_1, a_2, \cdots, a_n$ and $b_1, b_2, \cdots, b_n$ are sorted in the same way and minimal if the two sequences are sorted oppositely, one increasing and the other decreasing.

Compute $i^i$.

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 $$\sqrt{10+4\sqrt{3-2\sqrt{2}}}$$

The number $21982145917308330487013369$ is the thirteenth power of a positive integer. Which positive integer?

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 $$2^{\sqrt{2^{\sqrt{2^{\sqrt{2}^{\cdots}}}}}}$$

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 $$\sqrt{x\sqrt{x\sqrt{x\sqrt{\cdots}}}}=x$$

Compute $$\sqrt{\frac{2}{2^2}+\sqrt{\frac{2}{2^4}+\sqrt{\frac{2}{2^8}+\cdots}}}$$

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 $$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}}\approx 1$$


Simplify $\sqrt{5\sqrt{3}+6\sqrt{2}}$.

Simplify $\sqrt{12+2\sqrt{6}+2\sqrt{14}+2\sqrt{21}}$

Simplify $\sqrt{\sqrt[3]{9}+6\sqrt[3]{3}+9}$

Simplify $\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}$.

Solve $x^2 +6x - 4\sqrt{5}=0$.

Simplify $\sqrt{4+\sqrt[3]{81}+4\sqrt[3]{9}}$

Simplify $\sqrt{6+\sqrt[3]{81}+\sqrt[3]{9}}$.

A sequence satisfies $a_1 = 3, a_2 = 5$, and $a_{n+2} = a_{n+1} - a_n$ for $n \ge 1$. What is the value of $a_{2018}$?