展开 \exists x(x+2=1)

解答

展开

\exists x (x + 2 = 1)

\exists x^{2} + 2\exists x = 1

求解步骤

\exists x (x + 2 = 1)

= \exists x (x + 2 \cdot 1 =)

使用分配律:

a (b + c) = ab + a c

a = \exists x, b = x, c = 2 = 1

\exists xx + \exists x \cdot 2 = 1

\exists xx + 2\exists x = 1

乘开

\exists xx + 2\exists x : \exists x^{2} + 2\exists x

\exists x^{2} + 2\exists x = 1

e x p an d 4 (3 x + 2)^{2} + 4

e x p an d \frac{( 7 y - 6 ) ^{2}}{2}

e x p an d (a + a^{^{'}}) b

e x p an d \frac{2}{9} (3 e^{3 x} x - e^{3 x}) + C

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