Question

Найдите значение выражения 2(t+2)^2-(-t-2)^2+(2-t)(t+2) при t=-1/6

Answer 1

2(t^2+2t+4)-(t^2+2t+4)+4-t^2=2t^2+4t+8-t^2-2t-4+4-t^2=2t+8=2*(-1/6)+8=23/3

Answer 2

2 ( t + 2 )² - ( -t - 2 )² + ( 2 - t ) ( t + 2 ) при t = -1/6 :

2 ( t² + 4t + 4 ) - (t² + 4t + 4 ) + ( 2t + 4 - t² - 2t ) = 2t² + 8t + 8 - t² - 4t - 4 + 2t + 4 - t² - 2t = 4t + 8 = 4 × (-1/6) + 8 = (-⅔) + 8 =

$\frac{ - 2 + 24}{3} = \frac{22}{3} = 7 \frac{1}{3} .$

Ответ : 7⅓.

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