Question

Find the value of:
(i) (2⁰ + 3⁻¹) × 3²
(ii) (2⁻¹ × 3⁻¹) ÷ 2⁻³
(iii) ${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}$

Answer 1

(i)
​$$\begin{gathered} \left(2^0+3^{-1}\right)\times 3^2=\left(1+\frac{1}{3}\right)\times 3^2(\mathrm{because}{2}^0=1\mathrm{and}{3}^{-1}=\frac{1}{3}) \\ =\left(\frac{1\times 3}{1\times 3}+\frac{1\times 1}{3\times 1}\right)\times 3^2=\left(\frac{3}{3}+\frac{1}{3}\right)\times 3^2=\left(\frac{4}{3}\right)\times 3^2=4\times 3^{(2-1)}=4\times 3=12 \end{gathered}$$

(ii)

$$\begin{gathered} \left(2^{-1}\times 3^{-1}\right)÷2^{-3}=\left(\frac{1}{2}\times \frac{1}{3}\right)÷{\left(\frac{1}{2}\right)}^3 \\ \left(\frac{1}{6}\right)÷\frac{1^3}{2^3}=\left(\frac{1}{6}\right)÷\left(\frac{1}{8}\right)=\frac{1}{6}\times 8=\frac{8}{6}=\frac{4}{3} \end{gathered}$$


(iii)

${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}={\left(\frac{2}{1}\right)}^2+{\left(\frac{3}{1}\right)}^2+{\left(\frac{4}{1}\right)}^2=2^2+3^2+4^2=4+9+16=29$