Question

Find the value of:
(i) $(3^0+4^{-1})\times 2^2$

(ii) $(2^{-1}\times 4^{-1})÷2^{-2}$

(iii) ${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}$

(iv) $(3^{-1}+4^{-1}+5^{-1}{)}^0$

(v) ${\left[{\left(\frac{-2}{3}\right)}^{-2}\right]}^2$

(vi) $7^{-20}-7^{-21}$

Answer 1

(i) $(3^0+4^{-1})\times 2^2$$=(1+\frac{1}{4})\times 4$

$=\left(\frac{5}{4}\right)\times 4$

$=5$

(ii) $(2^{-1}\times 4^{-1})÷2^{-2}$$=(\frac{1}{2}\times \frac{1}{4})\times 2^2$

$=(\frac{1}{2}\times \frac{1}{4})\times 4$

$=\frac{1}{2}$

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

$=4+9+16$

$=29$

(iv) $(3^{-1}+4^{-1}+5^{-1}{)}^0$$={(\frac{1}{3}+\frac{1}{4}+\frac{1}{5})}^0$

$=1$ $\because a^0=1$

(v) ${\left[{\left(\frac{-2}{3}\right)}^{-2}\right]}^2$$={\left(\frac{-2}{3}\right)}^{-4}$

$={\left(\frac{-3}{2}\right)}^4$

$=\frac{81}{16}$

(vi) $7^{-20}-7^{-21}$$=\frac{1}{7^{20}}-\frac{1}{7^{21}}$

$=\frac{1}{7^{20}}(1-\frac{1}{7})$

$=\frac{1}{7^{20}}\left(\frac{6}{7}\right)$

$=\frac{6}{7^{21}}$