• # question_answer 3)                 Find the value of:                 (i) $({{3}^{0}}+{{4}^{-1}})\,\times {{2}^{2}}$                                      (ii) $({{2}^{-1}}\times {{4}^{-1}})\div {{2}^{-2}}$                 (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) ${{\left\{ {{\left( \frac{-2}{3} \right)}^{-2}} \right\}}^{2}}$

(i) $({{3}^{0}}+{{4}^{-1}})\,\times {{2}^{2}}$                 $({{3}^{0}}+{{4}^{-1}})\,\times {{2}^{2}}$$=\left( 1+\frac{1}{4} \right)\times 4$                 $=\frac{5}{4}\times 4$                 = 5                 (ii) $({{2}^{-1}}\times {{4}^{-1}})\div {{2}^{-2}}$                 $({{2}^{-1}}\times {{4}^{-1}})\div {{2}^{-2}}$$=\{{{2}^{-1}}\,\times {{({{2}^{2}})}^{-1}}\}\div {{2}^{-2}}$ $=\{{{2}^{-1}}\times {{2}^{2\,\times \,(-1)}}\}\,\div \,{{2}^{-2}}$ $=({{2}^{-1}}\times {{2}^{-2}})\,\div {{2}^{-2}}$ $={{2}^{(-1)\,+(-2)}}\div \,{{2}^{-2}}$ $={{2}^{-3}}\div \,{{2}^{-2}}$ $=\frac{{{2}^{-3}}}{{{2}^{-2}}}$ $=\frac{1}{{{2}^{(-2)\,-(-3)}}}$ $=\frac{1}{{{2}^{-2+3}}}$ $=\frac{1}{{{2}^{1}}}$ $=\frac{1}{2}$                 (iii) ${{\left( \frac{1}{2} \right)}^{-2}}\,+{{\left( \frac{1}{3} \right)}^{-2}}\,+{{\left( \frac{1}{4} \right)}^{-2}}$                 ${{\left( \frac{1}{2} \right)}^{-2}}\,+{{\left( \frac{1}{3} \right)}^{-2}}\,+{{\left( \frac{1}{4} \right)}^{-2}}$ $=\frac{{{1}^{-2}}}{{{2}^{-2}}}\,+\frac{{{1}^{-2}}}{{{3}^{-2}}}\,+\frac{{{1}^{-2}}}{{{4}^{-2}}}$ $=\frac{{{2}^{2}}}{{{1}^{2}}}\,+\frac{{{3}^{2}}}{{{1}^{2}}}\,+\frac{{{4}^{2}}}{{{1}^{2}}}$ $=\frac{4}{1}\,+\frac{9}{1}+\frac{16}{1}$ $=4+9+16$ = 29                 (iv) ${{({{3}^{-1}}+{{4}^{-1}}\,+{{5}^{-1}})}^{0}}$                 ${{({{3}^{-1}}+{{4}^{-1}}\,+{{5}^{-1}})}^{0}}$$={{\left[ \frac{1}{3}+\frac{1}{4}+\frac{1}{5} \right]}^{0}}$ $={{\left( \frac{20+15+12}{60} \right)}^{0}}$ $={{\left( \frac{47}{60} \right)}^{0}}$ $=1$                 (v) ${{\left\{ {{\left( \frac{-2}{3} \right)}^{-2}} \right\}}^{2}}$ ${{\left\{ {{\left( \frac{-2}{3} \right)}^{-2}} \right\}}^{2}}$$={{\left( \frac{-2}{3} \right)}^{(-2)\times 2}}$ $={{\left( \frac{-2}{3} \right)}^{-4}}$ $=\frac{{{(-2)}^{-4}}}{{{(3)}^{-4}}}$ $=\frac{{{3}^{4}}}{{{(-2)}^{4}}}$ $=\frac{3\times 3\times 3\times 3}{(-2)\times (-2)\times (-2)\times (-2)}$ $=\frac{81}{16}$