计算 (x+3)(x+4)-(x-1)(x+2)
1个回答

(x+3)(x+4)-(x-1)(x+2)

=x^2+7x+12-x^2-x+2

=6x+14

(2x的平方+5x+1)(x+2)-(x+2)(x+1)

=(x+2)(x^2+5x+1-x-1)

=(x+2)(x^2+4x)

=x(x+2)(x+4)

a的4次方-(a-b)(a+b)(a的平方-b的平方)

=a^4-(a^2-b^2)(a^2-b^2)

=a^4-(a^2-b^2)^2

=[a^2-(a^2-b^2)][a^2+(a^2-b^2)]

=b^2(2a^2-b^2)

(x-1)的平方(x+1)的平方(x的平方+1)的平方

=[(x-1)(x+1)]^2(x^2+1)^2

=(x^2-1)^2(x^2+1)^2

=[(x^2-1)(x^2+1)]^2

=(x^4-1)^2

=x^8-2x^4+1

14又8分之1×13又8分之7

=(14+1/8)*(13+7/8)

=(14+1/8)*(14-1/8)

=14^2-(1/8)^2

=196-1/64

=195又1/64

(3x+2y-1)(3x+2y+1)

=(3x+2y)^2-1

=9x^2+12xy+4y^2-1