展开 (x^2+Δx^2-2)^3

解答

展开

(x^{2} + Δ x^{2} - 2)^{3}

Δ^{3} x^{6} + 3 Δ^{2} x^{6} + 3Δ x^{6} + x^{6} - 6 Δ^{2} x^{4} - 12Δ x^{4} - 6 x^{4} + 12Δ x^{2} + 12 x^{2} - 8

求解步骤

(x^{2} + Δ x^{2} - 2)^{3}

(x^{2} + Δ x^{2} - 2)^{3} = (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2)

= (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2)

乘开

(x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2) : Δ^{3} x^{6} + 3 Δ^{2} x^{6} + 3Δ x^{6} + x^{6} - 6 Δ^{2} x^{4} - 12Δ x^{4} - 6 x^{4} + 12Δ x^{2} + 12 x^{2} - 8

= Δ^{3} x^{6} + 3 Δ^{2} x^{6} + 3Δ x^{6} + x^{6} - 6 Δ^{2} x^{4} - 12Δ x^{4} - 6 x^{4} + 12Δ x^{2} + 12 x^{2} - 8

e x p an d - 4 (x + Δ x)^{2}

e x p an d (1 + e^{4 x})^{2}

e x p an d - \frac{3 x}{h ^{2} ( x + h )}

e x p an d T (3, - 4, 5)

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