展开 t(t+dt+4)

解答

展开

t (t + d t + 4)

t^{2} + t d t + 4 t

求解步骤

t (t + d t + 4)

打开括号

= tt + t d t + t \cdot 4

= tt + t d t + 4 t

tt = t^{2}

= t^{2} + t d t + 4 t

e x p an d \frac{d}{d x} ((2 x^{2} - 1) e^{- x^{2}}) + 5

f (x) = \frac{( 5 x ^{5} + 10 x ^{2} + 1 )}{( 1 - x ^{2} ) ^{5}}

e x p an d \frac{1}{s ^{2} \cdot ( s + 4 ) ^{2}}

e x p an d - 2 (3 - 1) (1 + 2)

e x p an d (7 x^{3} + 9 y^{2} + 7 x y) d x