Write the following in the form $a\sqrt{b}$, where $b$ is a prime number:  - $\sqrt{300}-\sqrt{48}$
- $\sqrt{37.5}$
     - $6\sqrt{3}$
- $\frac{5}{2}\sqrt{6}$
       Write the following in the form $a+b\sqrt{c}$:  - $(7+\sqrt{5})(3-\sqrt{5})$
- $\dfrac{\sqrt{2}}{1-\sqrt{2}}$
     - $16-4\sqrt{5}$
- $-\sqrt{2}-2$
       Simplify  - $\dfrac{\sqrt{48}-\sqrt{600}}{\sqrt{12}}$
- $\dfrac{\sqrt{2}}{4+3\sqrt{2}}$
     - $2-5\sqrt{2}$
- $3-2\sqrt{2}$
       Simplify  - $\sqrt{48}-\dfrac{6}{\sqrt{3}}$
- $\dfrac{12\sqrt{3}}{\sqrt{50}-\sqrt{18}}$
        Simplify  - $\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}$
- $\dfrac{(32x^3)^{\frac{2}{5}}}{2x\sqrt{x}}$
     - $\dfrac{2}{5x^2}$
- $2x^{-\frac{3}{10}}$
       Solve $32\sqrt{2}=2^x$.     
  $x = \dfrac{11}{2}$   
      Given that $(2+\sqrt{5})x=6-\sqrt{5}$, find $x$ in the form $a+b\sqrt{5}$.     
  $8\sqrt{5}-17$   
      Solve:  - $10+x\sqrt{8}=\dfrac{6x}{\sqrt{2}}$
- $5^x\times5^{x+4}=25$
        Given that $y = \dfrac{1}{64}x^3$, find $4y^{-1}$ in the form $kx^n$.     
  $256x^{-3}$   
      Write:  - $2^6\times2^2$ in the form $8^n$
- $5\times4^{\frac{2}{3}}+3\times16^{\frac{1}{3}}$ in the form $2^n$
     - $8^{\frac{8}{3}}$
- $2^{\frac{13}{3}}$
       Find the integer $n$ such that $$n<5\sqrt{3}<n+1$$     
  $n=8$   
      The area of a rectangle of width $3\sqrt{2}-3$ is 6. Find the height of the rectangle.     
  $2\sqrt{2}+2$   
      Show that $\left(\dfrac{4}{3}\right)^{\frac{1}{2}}+\left(\dfrac{1}{3}\right)^{-\frac{1}{2}}$ can be written in the form $\dfrac{a}{b}\sqrt{c}$.     
  $\dfrac{5}{3}\sqrt{3}$   
      Given that $y=2^x$, express the following in terms of $y$:  - $2^{2x-1}$
- $8^x-4^{-x}$
     - $\dfrac{y^2}{2}$
- $y^3 - \dfrac{1}{y^2}$
       Write the following in the form $2^n$:  - $\Big(\big(2^3\big)^2\Big)^3$
- $\big(2^3\big)^{\big(2^3\big)}$
- $2^{\Big(\big(3^2\big)^3\Big)}$
- $2^{\Big(3^{\big(2^3\big)}\Big)}$
        A triangle $ABC$ has $AB=BC=4+\sqrt{3}$ and $AC=4+4\sqrt{3}$. $M$ is the midpoint of $AC$. Find the exact length of $BM$ and hence find the area of the triangle.     
  $BM=\sqrt{3}$, area $=6+2\sqrt{3}$   
      By first expanding $(1-\sqrt{5})^2$, solve $y^2=3-\sqrt{5}$.     
  $y = \pm \dfrac{\sqrt{2}-\sqrt{10}}{2}$   
      A triangle $ABC$ has $AB = 2\sqrt{3}-1$, $BC = \sqrt{3}+2$, and $\angle ABC = 90^{\circ}$.  - Find the area of the triangle.
- Show that $AC = 2\sqrt{5}$.
- Find the exact value of $\tan(\angle ACB)$.
     - $2+\dfrac{3\sqrt{3}}{2}$
- Algebra - see video
- $\tan(\angle ACB)=5\sqrt{3}-8$
       Solve $$25^{2x-1}=0.2^{x+3}$$     
  $x = -\frac{1}{5}$   
      Simplify $$\sqrt{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}$$     
  $3+2\sqrt{2}$   
      Evaluate $$\dfrac{1}{\sqrt{1}+\sqrt{2}} + \dfrac{1}{\sqrt{2}+\sqrt{3}} + ... + \dfrac{1}{\sqrt{24}+\sqrt{25}}$$     
  $4$   
      Given that $(a+b\sqrt{2})^2=59+30\sqrt{2}$, and that $a$ and $b$ are both positive, find $a$ and $b$.     
  $a = 3$ and $b = 5$   
        - Express $\sqrt[3]{24}$ in the form $a\sqrt[3]{3}$.
- Hence find the integer $n$ such that $$\sqrt[3]{n} =  \sqrt[3]{24}+\sqrt[3]{81}$$
     - $\sqrt[3]{8}\times\sqrt[3]{3} = 2\sqrt[3]{3}$
- $375$
       By considering $(a+b\sqrt{2})^2$, find $\sqrt{17-12\sqrt{2}}$.     
  $3-2\sqrt{2}$