Question

Evaluate :

$\left(i\right)(-5{)}^{\circ }(ii)8^{\circ }+4^{\circ }+2^{\circ }(iii\left)\right(8+4+2{)}^{\circ }\left(iv\right)4x^{\circ }\left(v\right)\left(4x{)}^{\circ }\right(vi\left)\right[\left({10}^3{)}^{\circ }{]}^5\right(vii\left)\right(7x^{\circ }{)}^2\left(viii\right)9^{\circ }+9^{-1}-9^{-2}+9^{\frac{1}{2}}-9^{-\frac{1}{2}}$

Answer 1

We know that $a^{\circ }=1$

$\left(i\right)(-5{)}^{\circ }=1(\because a^{\circ }=1\left)\right(ii)8^{\circ }+4^{\circ }+2^{\circ }=1+1+1=3(\because a^{\circ }=1\left)\right(iii\left)\right(8+4+2{)}^{\circ }=(14{)}^{\circ }=1(\because a^{\circ =1}\left)\right(iv)4x^{\circ }=4\times 1=4(v\left)\right(4x{)}^{\circ }=1\left(vi\right)\left[\right({10}^3{)}^{\circ }{]}^5={10}^{\circ }=1\left(vii\right)(7x^{\circ }{)}^2=7^2\times x^{0\times 2}=49\times 1=49(viii)9^{\circ }+9^{-1}-9^{-2}+9^{\frac{1}{2}}-9^{-\frac{1}{2}}=1+\frac{1}{9}-\frac{1}{9^2}=(3^2{)}^{\frac{1}{2}}-(3^2{)}^{-\frac{1}{2}}=1+\frac{1}{9}-\frac{1}{81}+3^{2\times \frac{1}{2}}-3^{2\times (-\frac{1}{2})}=1+\frac{1}{9}-\frac{1}{81}+3-3^{-1}=1+\frac{1}{9}-\frac{1}{81}+\frac{3}{1}-\frac{1}{3}=\frac{81+9-1+243-27}{81}=\frac{333-28}{81}=\frac{305}{81}=3\frac{62}{81}$