\(a,9.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{7}\right)^0-1\\ =9.\dfrac{1}{9}+1-1\\ =\dfrac{9}{9}+1-1\\ =1+1-1\\ =1\\ b,7.\left(5^2-5\right)+\left(-1\right)^2\\ =7.\left(25-5\right)+1\\ =7.20+1\\ =140+1\\ =141\)
a: =9*1/9+1-1=1
b: =7*(25-5)-1=7*20-1=139
`a, 9 xx (1/3)^2 + (1/7)^0 - 1`
`= 9 xx 1/9 + 1 -1`
`= 1`
`b, 7(5^2-5) + (-1)^2`
`= 7 . 20 + 1`
`= 141`
`a)` `9.(1/3)^2+(1/7)^(o)-1`
`=9.(1)/(9)+1-1`
`=1`
`b)` `7.(5^(2)-5)+(-1)^2`
`=7.20+1`
`=140+1`
`=141`