展开 x(1+Δ)^2

解答

展开

x (1 + Δ)^{2}

x + 2Δ x + Δ^{2} x

求解步骤

x (1 + Δ)^{2}

(1 + Δ)^{2} = 1 + 2Δ + Δ^{2}

= (Δ^{2} + 2Δ + 1) x

= (1 + 2Δ + Δ^{2}) x

打开括号

= x \cdot 1 + x \cdot 2Δ + x Δ^{2}

= 1 \cdot x + 2Δ x + Δ^{2} x

乘以:

1 \cdot x = x

= x + 2Δ x + Δ^{2} x

\frac{d}{d y} ((x) (x y + sin (x y) + x))

e x p an d \frac{( 4 x - 3 ) ( x + 2 )}{8 x + 6}

e x p an d 1 (- 4 x + 2 y = 6)

e x p an d \frac{( 2 x - 1 )}{( 3 x - 1 ) ( x - 1 )}

e x p an d \frac{5 ( 3 30 - 10 )}{34}