展开 α(cα,aα+bα+cα,4aα+2bα+cα)

解答

展开

α (c α, a α + b α + c α, 4 a α + 2 b α + c α)

αα c, α a + αα b + αα c, 4 α a + 2 αα b + αα c

求解步骤

α (c α, a α + b α + c α, 4 a α + 2 b α + c α)

= α (, αα c a + α b + 4, αα c a + 2 α b + α c)

打开括号

= α c α, a α + α b α + α c α, 4 a α + α \cdot 2 b α + α c α

= αα c, α a + αα b + αα c, 4 α a + 2 αα b + αα c

e x p an d - \frac{8 s + 25}{( s + 4 ) ^{2}}

e x p an d (1 + j) e^{\frac{jπ t}{2}}

e x p an d (p + t) (e + t)

e x p an d \frac{100 ( s + 25 )}{s ( s + 5 ) ^{3}}

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