Question

急！完全平方公式计算（超简单）要过程

急！完全平方公式计算（超简单）要过程
(3x+0.5)(0.5-3x)
(a-b+c)(a-b-c)
(2x-3y/2)(3y+2x/3)
(2x-3y)^2(2x+3y)^2
(5a^2-b^4)^2
1.02x0.98
1又2/13x11/13
(40又1/2)^2
2005^2-4010x2006+2006^2
还有平方差公式

Answer 1

(3x+0.5)(0.5-3x)=(3x)^2-0.5^2=9x^2-0.25
(a-b+c)(a-b-c)=[(a-b)+c][(a-b)-c]=(a-b)^2-c^2=a^2-2ab+b^2-c^2
(2x-3y/2)(3y+2x/3)=4xy+4/3x^2-9/2y^2
(2x-3y)^2(2x+3y)^2=[(2x)^2-(3y)^2]^2=16x^4-72x^2y^2+81y^4
(5a^2-b^4)^2=25a^4-10a^2b^4+b^16
1.02x0.98(1+0.02)(1-0.02)=1-0.0004=0.9996
1又2/13x11/13=15/13x11/13=165/169
(40又1/2)^2=(81/2)^2=6561/4
2005^2-4010x2006+2006^2=(2006-2005)^2=1
赞一个

Answer 2

(3x+0.5)(0.5-3x)=(3x)^2-0.5^2=9x^2-0.25
(a-b+c)(a-b-c)=[(a-b)+c][(a-b)-c]=(a-b)^2-c^2=a^2-2ab+b^2-c^2
(2x-3y/2)(3y+2x/3)=4xy+4/3x^2-9/2y^2
(2x-3y)^2(2x+3y)^2=[(2x)^2-(3y)^2]^2=16x^4-72x^2y^2+81y^4
(5a^2-b^4)^2=25a^4-10a^2b^4+b^16
1.02x0.98(1+0.02)(1-0.02)=1-0.0004=0.9996
1又2/13x11/13=15/13x11/13=165/169
(40又1/2)^2=(81/2)^2=6561/4
2005^2-4010x2006+2006^2=(2006-2005)^2=1

Answer 3

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