Question

(7a−11b)⋅(7a+11b)−49a2, если a=3 и b=0,01

Answer 1

(7 × 3 - 11 × 0,01) × (7 × 3 + 11 × 0,01) - 49 × 3²

(21 - 0,11) × (21 + 0,11) - 49 × 9

20,89 × 21,11 - 441

440,9879 - 441

- 0,0121

Answer 2

$(7a-11b)(7a+11b)-49a^2=49a^2-121b^2-49a^2=-121b^2\\\\-121*(0,01)^2=-121*(\frac{1}{100} )^2=-\frac{121}{10000} =-0,0121$

или

$-121*(0,01)^2=-121*(10^{-2})^2=-121*10^{-4}=-0,0121$