\({\left({x}^{3}\right)}^{\frac{4}{3}}\)
\[\left( x^3 \right)^{\frac{4}{3}} = x^4\]
\({\left({s}^{2}\right)}^{\frac{1}{2}}\)
\[\left( s^2 \right)^{\frac{1}{2}} = s\]
\({\left({m}^{5}\right)}^{\frac{5}{3}}\)
\[\left( m^5 \right)^{\frac{5}{3}} = m^{\frac{25}{3}}\]
\({\left(-{m}^{2}\right)}^{\frac{4}{3}}\)
\[\left(- m^2 \right)^{\frac{4}{3}} = m^{\frac{8}{3}}\]
\(-{\left({m}^{2}\right)}^{\frac{4}{3}}\)
\[-\left( m^2 \right)^{\frac{4}{3}} = -m^{\frac{8}{3}}\]
\({\left(3{y}^{\frac{4}{3}}\right)}^{4}\)
\[\left( 3y^{\frac{4}{3}} \right)^4 = 81y^{\frac{16}{3}}\]
\(\dfrac{3{a}^{-2}{b}^{15}{c}^{-5}}{{\left({a}^{-4}{b}^{3}c\right)}^{\frac{-5}{2}}}\)
\begin{align*}
\dfrac{3a^{-2}b^{15}c^{-5}}{(a^{-4}b^3c)^{-\frac{5}{2}}} &=
\dfrac{3a^{-2}b^{15}c^{-5}}{a^{10}b^{-\frac{15}{2}}c^{-\frac{5}{2}}} \\
&= \dfrac{3b^{15+\frac{15}{2}}}{a^{12}c^{5-\frac{5}{2}}} \\
&= \dfrac{3b^{\frac{45}{2}}}{a^{12}c^{\frac{5}{2}}}
\end{align*}
\({\left(9{a}^{6}{b}^{4}\right)}^{\frac{1}{2}}\)
\[\left( 9a^6b^4 \right)^{\frac{1}{2}} = 3a^3b^2\]
\({\left({a}^{\frac{3}{2}}{b}^{\frac{3}{4}}\right)}^{16}\)
\[\left( a^{\frac{3}{2}}b^{\frac{3}{4}} \right)^{16} = a^{24}b^{12}\]
\({x}^{3}\sqrt{x}\)
\begin{align*}
x^3 \sqrt{x} &= x^3 \times x^{\frac{1}{2}} \\
&= x^{\frac{6}{2}} \times x^{\frac{1}{2}} \\
&= x^{\frac{7}{2}}
\end{align*}
\(\sqrt[3]{{x}^{4}{b}^{5}}\)
\begin{align*}
\sqrt[3]{x^4b^5} &= \left( x^4b^5 \right)^{\frac{1}{3}} \\
&= x^{\frac{4}{3}}b^{\frac{5}{3}}
\end{align*}
Re-write the following expression as a power of \(x\):
\(\dfrac{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}}{x^2}\)
\begin{align*}
\dfrac{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}}{x^2} &= \dfrac{x \times
x^{\frac{1}{2}} \times x^{\frac{1}{4}} \times x^{\frac{1}{8}} \times
x^{\frac{1}{16}} }{x^2} \\
&= \dfrac{x^{\frac{31}{16}} }{x^{\frac{32}{16}}} \\
&= \dfrac{1}{x^{\frac{1}{16}}}
\end{align*}
Expand:
\(\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)\)
\begin{align*}
\left( \sqrt{x} - \sqrt{2} \right) \left( \sqrt{x} + \sqrt{2} \right) &=
\left( \sqrt{x} \right)^2 - \left( \sqrt{2} \right)^2 \\
&= x - 2
\end{align*}
Rationalise the denominator:
\(\dfrac{10}{\sqrt{x}-1}\)
\begin{align*}
\dfrac{10}{\sqrt{x} - 1} &= \dfrac{10}{\sqrt{x} - 1} \times \dfrac{\sqrt{x}
+ 1}{\sqrt{x} + 1} \\
&= \dfrac{10 \left( \sqrt{x} + 1 \right) }{\left( \sqrt{x} \right)^2 - 1} \\
&= \dfrac{10 \sqrt{x} + 10}{x-1}
\end{align*}
Write as a single term with a rational denominator:
\(\dfrac{3}{2\sqrt{x}}+\sqrt{x}\)
\begin{align*}
\dfrac{3}{2\sqrt{x}} + \sqrt{x} &= \dfrac{3 + 2\sqrt{x}\sqrt{x}}{2\sqrt{x}}
\\
&= \dfrac{3 + 2x}{2\sqrt{x}} \times \frac{\sqrt{x} }{\sqrt{x} } \\
&= \dfrac{3\sqrt{x} + 2x\sqrt{x} }{2x}
\end{align*}
\begin{align*}
\sqrt{72} &= \sqrt{8 \times 9} \\
&= \sqrt{8} \times \sqrt{9} \\
&= 2\sqrt{2} \times 3 \\
&= 6\sqrt{2}
\end{align*}
\begin{align*}
\sqrt{45} + \sqrt{80} &= \sqrt{5 \times 9} + \sqrt{5 \times 16} \\
&= 3\sqrt{5} + 4\sqrt{5} \\
&= 7\sqrt{5}
\end{align*}
\(\dfrac{\sqrt{48}}{\sqrt{12}}\)
\begin{align*}
\dfrac{\sqrt{48}}{\sqrt{12}} &= \sqrt{\dfrac{48}{12}} \\
&= \sqrt{4} \\
&= 2 \\
\text{ or }
\dfrac{\sqrt{48}}{\sqrt{12}} &= \dfrac{\sqrt{3 \times 16}}{\sqrt{3
\times 4}} \\
&= \dfrac{4\sqrt{3}}{2\sqrt{3}} \\
&= \frac{4}{2} \\
&= 2
\end{align*}
\(\dfrac{\sqrt{18}÷\sqrt{72}}{\sqrt{8}}\)
\begin{align*}
\dfrac{\sqrt{18} \div \sqrt{72}}{\sqrt{8}} &=
\dfrac{\sqrt{\frac{18}{72}}}{\sqrt{8}} \\
&= \dfrac{\sqrt{\frac{1}{4}}}{2\sqrt{2}} \\
&= \dfrac{\frac{1}{2}}{2\sqrt{2}} \\
&= \dfrac{1}{2} \times \frac{1}{2\sqrt{2}} \\
&= \dfrac{1}{4\sqrt{2}}
\end{align*}
\(\dfrac{4}{\left(\sqrt{8}÷\sqrt{2}\right)}\)
\begin{align*}
\dfrac{4}{\sqrt{8} \div \sqrt{2}} &= \dfrac{4}{2\sqrt{2} \div
\sqrt{2}} \\
&= \dfrac{4}{2} \\
&= 2
\end{align*}
\(\dfrac{16}{\left(\sqrt{20}÷\sqrt{12}\right)}\)
\begin{align*}
\dfrac{16}{\sqrt{20} \div \sqrt{12}} &= \dfrac{16}{2\sqrt{5} \div
2\sqrt{3}} \\
&= \dfrac{16}{\frac{2\sqrt{5}}{2\sqrt{3}}} \\
&= 16 \times \dfrac{\sqrt{3}}{\sqrt{5}} \\
&= 16 \times \dfrac{\sqrt{3}}{\sqrt{5}} \times
\dfrac{\sqrt{5}}{\sqrt{5}} \\
&= \dfrac{16\sqrt{15}}{5}
\end{align*}
\({\left(2+\sqrt{2}\right)}^{2}\)
\begin{align*}
\left( 2 + \sqrt{2} \right)^2 &= \left( 2 + \sqrt{2} \right) \left(
2 + \sqrt{2} \right) \\
&= 4 + 4\sqrt{2} + \left( \sqrt{2} \right)^2 \\
&= 4 + 4\sqrt{2} + 2 \\
&= 6 + 4\sqrt{2}
\end{align*}
\(\left(2+\sqrt{2}\right)\left(1+\sqrt{8}\right)\)
\begin{align*}
\left( 2 + \sqrt{2} \right) \left( 1 + \sqrt{8} \right) &= \left( 2
+ \sqrt{2} \right) \left( 1 + 2\sqrt{2} \right) \\
&= 2 + 4\sqrt{2} + \sqrt{2} + 2\left( \sqrt{2} \right)^2 \\
&= 2 + 5\sqrt{2} + 2(2) \\
&= 6 + 5\sqrt{2}
\end{align*}
\(\left(1+\sqrt{3}\right)\left(1+\sqrt{8}+\sqrt{3}\right)\)
\begin{align*}
\left( 1 + \sqrt{3} \right) \left( 1 + \sqrt{8} + \sqrt{3} \right)
&= 1 + \sqrt{8} + \sqrt{3} + \sqrt{3} + \sqrt{8 \times 3} + \left(
\sqrt{3} \right)^2\\
&= 1 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6} + 3 \\
&= 4 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}
\end{align*}
\(\sqrt{5}\left(\sqrt{45}+2\sqrt{80}\right)\)
\begin{align*}
\sqrt{5} \left( \sqrt{45} + 2\sqrt{80} \right) &= \sqrt{5} \left(
\sqrt{9 \times 5} + 2\sqrt{16 \times 5} \right) \\
&= \sqrt{5} \left( 3\sqrt{5} + 8\sqrt{5} \right) \\
&= 3 \times 5 + 8 \times 5 \\
&= 15 + 40 \\
&= 55
\end{align*}
\(\dfrac{\sqrt{98}-\sqrt{8}}{\sqrt{50}}\)
\begin{align*}
\dfrac{\sqrt{98} - \sqrt{8}}{\sqrt{50}} &= \dfrac{\sqrt{49 \times 2}
- \sqrt{4 \times 2}}{\sqrt{25 \times 2}} \\
&= \dfrac{7\sqrt{2} - 2\sqrt{2}}{5\sqrt{2}} \\
&= \dfrac{5\sqrt{2}}{5\sqrt{2}} \\
&= 1
\end{align*}
Simplify:
\(\sqrt{98{x}^{6}}+\sqrt{128{x}^{6}}\)
\begin{align*}
\sqrt{98x^6} + \sqrt{128x^6} &= \sqrt{2 \times 49 x^6} + \sqrt{2 \times
64x^6} \\
&= 7x^3 \sqrt{2} + 8x^3 \sqrt{2} \\
&= 15 \sqrt{2}x^3
\end{align*}
\(\dfrac{\sqrt{5}+2}{\sqrt{5}}\)
\begin{align*}
\dfrac{\sqrt{5} + 2 }{\sqrt{5}} &= \dfrac{\sqrt{5} + 2 }{\sqrt{5}}
\times \dfrac{\sqrt{5}}{\sqrt{5}} \\
&= \dfrac{ 5 + 2\sqrt{5} }{5} \\
&= 1 + \dfrac{2\sqrt{5}}{5}
\end{align*}
\(\dfrac{y-4}{\sqrt{y}-2}\)
\begin{align*}
\dfrac{\sqrt{y} - 4 }{\sqrt{y} - 2 } &= \dfrac{y - 4 }{\sqrt{y} - 2
} \times \dfrac{\sqrt{y} + 2}{\sqrt{y} + 2} \\
&= \dfrac{y\sqrt{y} +2y - 4\sqrt{y} - 8}{y - 4} \\
&= \dfrac{2y + y\sqrt{y} - 4\sqrt{y} - 8}{y - 4}
\end{align*}
\(\dfrac{2x-20}{\sqrt{x}-\sqrt{10}}\)
\begin{align*}
\dfrac{2x - 20 }{\sqrt{x} - \sqrt{10} } &= \dfrac{2x - 20 }{\sqrt{x}
- \sqrt{10} } \times \dfrac{\sqrt{x} + \sqrt{10}}{\sqrt{x} + \sqrt{10}}
\\
&= \dfrac{2(x - 10) \left( \sqrt{x} + \sqrt{10} \right) }{x - 10 }
\\
&= 2\sqrt{x} + 2\sqrt{10}
\end{align*}
Evaluate without using a calculator:
\({\left(2-\dfrac{\sqrt{7}}{2}\right)}^{\frac{1}{2}} \times
{\left(2+\dfrac{\sqrt{7}}{2}\right)}^{\frac{1}{2}}\)
\begin{align*}
\left( 2 - \dfrac{\sqrt{7}}{2} \right)^{\frac{1}{2}} \left( 2 +
\dfrac{\sqrt{7}}{2} \right)^{\frac{1}{2}} &= \left( 4 - \frac{7}{4}
\right)^{\frac{1}{2}} \\
&= \sqrt{\frac{16}{4} - \frac{7}{4}}\\
&= \sqrt{\frac{9}{4}}\\
&= \frac{3}{2}
\end{align*}
Prove (without the use of a calculator):
\(\sqrt{\dfrac{8}{3}}+5\sqrt{\dfrac{5}{3}}-\sqrt{\dfrac{1}{6}}=\dfrac{10\sqrt{15}+3\sqrt{6}}{6}\)
\begin{align*}
\text{LHS } &= \sqrt{\dfrac{8}{3}} + 5 \sqrt{\dfrac{5}{3}} -
\sqrt{\dfrac{1}{6}} \\
&= \sqrt{\frac{8}{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} + 5
\sqrt{\dfrac{5}{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} - \sqrt{\dfrac{1}{6}}
\times \dfrac{\sqrt{6}}{\sqrt{6}} \\
&= \dfrac{\sqrt{8}\sqrt{3}}{3} + 5\frac{\sqrt{5}\sqrt{3}}{3} -
\frac{\sqrt{6}}{6} \\
&= \dfrac{ 2\sqrt{24}}{6} + \frac{10\sqrt{15}}{6} - \frac{\sqrt{6}}{6} \\
&= \dfrac{ 4\sqrt{6} + 10\sqrt{15} - \sqrt{6}}{6} \\
&= \dfrac{ 10\sqrt{15} + 3\sqrt{6}}{6} \\
&= \text{RHS }
\end{align*}
Simplify completely by showing all your steps (do not use a calculator):
\[{3}^{-\frac{1}{2}}\left[\sqrt{12}+\sqrt[3]{\left(3\sqrt{3}\right)}\right]\]
\begin{align*}
3^{-\frac{1}{2}} \left( \sqrt{12} + \sqrt[3]{3\sqrt{3}} \right) &=
\frac{1}{\sqrt{3}} \left( 2\sqrt{3} + \sqrt[3]{3\times3^{\frac{1}{2}}} \right)
\\
&= \frac{1}{\sqrt{3}} \left( 2\sqrt{3} + \left( 3^{\frac{3}{2}}
\right)^{\frac{1}{3}} \right) \\
&= \frac{1}{\sqrt{3}} \left( 2\sqrt{3} + 3^{\frac{1}{2}} \right) \\
&= \frac{1}{\sqrt{3}} \left( 3\sqrt{3} \right) \\
&= 3
\end{align*}
Fill in the blank surd-form number on the right hand side of the equal sign
which will make the following a true statement: \(-3\sqrt{6}\times
-2\sqrt{24}=-\sqrt{18}\times ...\)
\begin{align*}
\text{LHS } &= -3\sqrt{6} \times -2\sqrt{24} \\
&= 6\sqrt{6}\sqrt{4 \times 6} \\
&= 12\sqrt{6}\sqrt{6} \\
&= 12(6) \\
&= 72 \\
\text{So then if LHS } &= \text{RHS } \\
\text{RHS } &= 72 \\
&= \sqrt{\text{5 184}} \\
&= -\sqrt{18} \times -\sqrt{\text{288}}
\end{align*}
\begin{align*}
3^{x - 1} - 27 &= 0\\
3^{x - 1} &= 27\\
3^{x - 1} &= 3^3\\
x - 1 &= 3 \\
\therefore x &= 4
\end{align*}
\(8^x - \frac{1}{\sqrt[3]{8}} = 0\)
\begin{align*}
8^x - \frac{1}{\sqrt[3]{8}} &= 0\\
2^{3x} &= \frac{1}{\sqrt[3]{2^3}} \\
2^{3x} &= \frac{1}{2} \\
2^{3x} &=2^{-1} \\
\therefore x &= -\frac{1}{3} \\
\text{ or } & \\
8^x - \frac{1}{\sqrt[3]{8}} &= 0\\
8^x &= \frac{1}{\sqrt[3]{8}} \\
8^x &= 8^{-\frac{1}{3}}\\
\therefore x &= -\frac{1}{3}
\end{align*}
\begin{align*}
27(4^x) &= (64)3^x \\
\frac{27}{64} &= \frac{3^x}{4^x} \\
\frac{3^3}{4^3} &= \left( \frac{3}{4} \right)^x \\
\left( \frac{3}{4} \right)^3 &= \left( \frac{3}{4} \right)^x \\
\therefore x &= 3
\end{align*}
\(\sqrt{2x - 5} = 2 - x\)
\begin{align*}
\sqrt{2x - 5} &= 2 - x \\
\left( \sqrt{2x - 5} \right)^2 &= \left( 2 - x \right)^2\\
2x - 5 &= 4 - 4x + x^2 \\
0 &= x^2 - 6x + 9 \\
0 &= (x - 3)(x - 3) \\
\therefore x &= 3 \\
\text{Check solution: LHS } &= \sqrt{2(3) - 5} \\
&= \sqrt{6 - 5} \\
&= \sqrt{1} \\
&= 1 \\
\text{Check solution: RHS } &= 2-3 \\
&= -1 \\
\text{RHS } &\ne \text{LHS }\\
\therefore &\text{No solution}
\end{align*}
\(2x^{\frac{2}{3}} + 3x^{\frac{1}{3}} - 2 = 0\)
\begin{align*}
2x^{\frac{2}{3}} + 3x^{\frac{1}{3}} - 2 &= 0 \\
\left( 2x^{\frac{1}{3}} - 1 \right) \left( x^{\frac{1}{3}} + 2 \right)
&= 0 \\
\therefore 2x^{\frac{1}{3}} - 1 &= 0 \\
2x^{\frac{1}{3}} &= 1 \\
x^{\frac{1}{3}} &= \frac{1}{2} \\
\left( x^{\frac{1}{3}} \right)^3 &= \left( \frac{1}{2} \right)^3 \\
\therefore x &= \frac{1}{8} \\
\text{ or } & \\
x^{\frac{1}{3}} + 2 &= 0 \\
x^{\frac{1}{3}} &= -2 \\
\left( x^{\frac{1}{3}} \right)^3 &= \left( -2 \right)^3 \\
x &= -8 \\
\text{Therefore } x = \frac{1}{8} &\text{ or } x = -8
\end{align*}
