(ab+1)^2-(a+b)^2因式分解

leon_chan2022-10-04 11:39:543条回答

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祸祸共回答了15个问题 | 采纳率93.3%: (ab+1)^2-(a+b)^2
=(ab+1+a+b)(ab+1-a-b)
=[a(b+1)+(b+1)][a(b-1)-(b-1)]
=(a+1)(b+1)(a-1)(b-1); 1年前

fechry 共回答了469个问题 | 采纳率: (ab+1)²-(a+b)²
=[(ab+1)+(a+b)][(ab+1)-(a+b)]
=[ab+a+b+1][ab-a+1-b]
=[a(b+1)+(b+1)][a(b-1)+(1-b)]
=(a+1)(b+1)(a-1)(b-1); 1年前

jin_shan 共回答了250个问题 | 采纳率: (ab+1)^2-(a+b)^2
(a^2b^2+1+2ab)-(a^2+2ab+b^2)
a^2b^2+1+2ab-a^2-2ab-b^2
a^2b^2-a^2-b^2+1
a^2(b^2-1)-(b^2-1)
(a^2-1)(b^2-1)
(a+1)(a-1)(b+1)(b-1); 1年前

分组分解法 1）9m^2-6m+2n-n^2 2）4x^2+1/4-9y^2-2x 3) (ab+1)^2-(a+b)^ 分组分解法 1）9m^2-6m+2n-n^2 2）4x^2+1/4-9y^2-2x 3) (ab+1)^2-(a+b)^2 4) (z^2-x^2-y^2)^2-4x^2y^2 honghao231年前1: 茅厕11 共回答了23个问题 | 采纳率91.3%

1）9m²-6m+2n-n²
=9m²-6m+1-(n²-2n+1)
=(3m-1)²-(n-1)²
=(3m-1+n-1)(3m-1-n+1)
=(3m+n-2)(3m-n)
2）4x²+1/4-9y²-2x
=4x²-2x+1/4-9y²
=(2x-1/2)²-9y²
=(2x-1/2 +3y)(2x-1/2 -3y)
=(1/4)(4x+6y-1)(4x-6y-1)
3) (ab+1)²-(a+b)²
=(ab+1+a+b)(ab+1-a-b)
=(a+1)(b+1)(1-a)(1-b)
4) (z²-x²-y²)²-4x²y²
=(z²-x²-y²+2xy)(z²-x²-y²-2xy)
=[z²-(x-y)²]*[z²-(x+y)²]
=(z+x-y)(z-x+y)(z+x+y)(z-x-y)

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简单的化简求值1.（ab+1)^2-(ab-1)^22.x(x+2y)-(x+1)^23Q（十万火急） 喑黑_cc1年前8: 呼吸不说谎_ 共回答了18个问题 | 采纳率100%

1.（ab+1)^2-(ab-1)^2
=[(ab+1)-(ab-1)][(ab+1)+(ab-1)]
=2*2ab
=4ab
2.x(x+2y)-(x+1)^2
=x^2+2xy-x^2-2x-1
=2xy-2x-1

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1.（ab+1)^2-(ab-1)^2 1.（ab+1)^2-(ab-1)^2 2.(3mn+1)(3mn-1)-8m^2n^2 3.(x+2)^2-(x+5)(x-1) 4.(x-2)^2-2(x+1)(x-1) 5.(x+2)^2-(3x+1)(x-1) 6.(-2a+10b)(2a+10b) lygang121年前4: jimmy0314 共回答了12个问题 | 采纳率91.7%

1.（ab+1)^2-(ab-1)^2
=(ab+1+ab-1)(ab+1-ab+1)=4ab
2.(3mn+1)(3mn-1)-8m^2n^2
=9m^2n^2-1-8m^2n^2
=m^2n^2-1
3.(x+2)^2-(x+5)(x-1)
=x^2+4x+4-x^2-4x+5
=9
4.(x-2)^2-2(x+1)(x-1)
=x^2-4x+4-2x^2+1
=-x^2-4x+5
=(x+5)(1-x)
5.(x+2)^2-(3x+1)(x-1)
=x^2+4x+4-3x^2+2x+1
=-2x^2+6x+5
6.(-2a+10b)(2a+10b)
=100b^2-4a^2

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七年级下册数学书45页知识技能要过程（x-2）（x+2）-（x+1)(x-3) (ab+1)^2-(ab-1)^2 (2 七年级下册数学书45页知识技能 要过程（x-2）（x+2）-（x+1)(x-3) (ab+1)^2-(ab-1)^2 (2x-y)^2-4(x-y)(x+2y) 麟台夜月1年前3: asdfgbvcxz163 共回答了16个问题 | 采纳率100%

x2-4-(x2-2x-3)=2x-1 a^2b^2+1-a^2b^2+1=2 4x^2-4xy+y^2-4x^2-8xy+4xy+8y^2=9y^2-8xy

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因式分解:(a^2b^2+ab)^2-(ab+1)^2 sweetwish20061年前1: Minanoto_G 共回答了20个问题 | 采纳率100%

(a^2b^2+ab)^2-(ab+1)^2 =[(a^2b^2+ab)+(ab+1)][(a^2b^2+ab)-(ab+1)] =(a^2b^2+2ab+1)(a^2b^2-1) =(ab+1)^2(ab+1)(ab-1) =(ab+1)^3(ab-1)

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