(2m-3n)-2(2m-3n)(-2n-3n) (2m 3n)2

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(2m-3n)-2(2m-3n)(-2n-3n) (2m 3n)2
计算(3-m+n)*(3+m-n)-(2-m+n)*(2+m-n)

原式=[3-(m-n)][3+(m-n)]-[2-(m-n)][2+(m-n)]=[3²-(m-n)²]-[2²-(m-n)²]=9-(m-n)²-4

A.3M+N B.2M+2N C.M+N D.M+3N

解题思路:考察长方形的性质:长方形的对边相等,长方形的周长等于四边之和解题过程:解:因为长方形的对边相等所以另一边的长=[4M-2(M-N)]/2=M+N选C最终答案:C

计算[5(m+n)2(m-n)]3÷【3(m+n)2(n-m)]2

[5(m+n)2(m-n)]3÷【3(m+n)2(n-m)]2=125(m+n)的6次方(m-n)³÷9(m+n)的4次方(m-n)²=125/9(m+n)²(m-n)如

计算(m+n)^3-(n+m)(m-n)^2

(m+n)^3-(n+m)(m-n)^2=(m+n)[(m+n)^2-(m-n)^2]=(m+n)[(m+n+m-n)(m+n-m+n)]=(m+n)(2m+2n)=2(m+n)^2

(m+5n)^2-2(5n+m)(n-3m)+(3m-n)^2

(m+5n)^2-2(5n+m)(n-3m)+(3m-n)^2=(m+5n)^2-2(m+5n)(n-3m)+(n-3m)^2=[(m+5n)-(n-3m)]^2=(4m+4n)^2=16(m-n)^

计算:5(m+n)(m-n)-2(m+n)^2-3(m-n)^2

5(m+n)(m-n)-2(m+n)^2-3(m-n)^2=5(m^2-n^2)-2(m^2+2mn+n^2)-3(m^2-2mn+n^2)=5m^2-5n^2-2m^2-4mn-2n^2-3m^2+

(3m-4n)(4n+3m)-(2m-n)(2m+n)

(3m-4n)(4n+3m)-(2m-n)(2m+n)=(9m²-16n²)-(4m²-n²)=9m²-16n²-4m²+n

(2m+3n)(2m-3n)-(3m-2n)(3m+2n)

(2m+3n)(2m-3n)-(3m-2n)(3m+2n)=(4m²-9n²)-(9m²-4n²)=4m²-9n²-9m²+4n&

(n-m)^3×(m-n)^2-(m-n)^5

(n-m)^3×(m-n)^2-(m-n)^5=-(m-n)^3*(m-n)^2-(m-n)^5=-(m-n)^5-(m-n)^5=-2(m-n)^5

(3m-2n)^+(3m+2n) (3m-2n)-9(m+n)^

(3m-2n)²+(3m+2n)(3m-2n)-9(m+n)²=(3m-2n)²-9(m+n)²+(3m+2n)(3m-2n)=(3m-2n+3m+3n)(3m

[(m-n)^2*(m-n)^3]^2/(m-n)^4

设m-n为a(a^2*a^3)^2/a^4=a^6即(m-n)^6

因式分解(n-m)^3(m-n)^2-(m-n)^5

解原式=-(m-n)³(m-n)²-(m-n)^5=-(m-n)^5-(m-n)^5=-2(m-n)^5

(2m+3n)(3m-2n)-(3m-2n)(3m+2n)

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已知3m=2n,则m/(m+n)+n/(m-n)-n^2/(m^2-n^2)=?

m/(m+n)+n/(m-n)-n^2/(m^2-n^2)=[m(m-n)+n(m+n)-n^2]/(m^2-n^2)=m^2/(m^2-n^2)=1/(1-(n/m)^2)=1/(1-(3/2)^2

已知m/n=5、,求(m/(m+n))+(m/(m-n))-(n^2/(m^3-n^2))

已知m=5n,则原式=(5n/(5n+n))+(5n/(5n-n))-(n^2)/(((5n)^3)-n^2)=(5/6)+(5/4)-[1/(125n-1)]=(25/12)-[1/(125n-1)

[2(m+n)^5-3(m+n)^4+(-m-n)^3]÷[2(m+n)^3]

原式=2(m+n)^5÷[2(m+n)^3]-3(m+n)^4÷[2(m+n)^3]+(-m-n)^3÷[2(m+n)^3]=(m+n)^2-3(m+n)/2-1/2再问:��Ҳ�㵽��һ���ˣ��

2n(m+n)^2+2m(m+n)^2+(m+n)^3 分解因式:

2n(m+n)^2+2m(m+n)^2+(m+n)^3=(m+n)[2n(m+n)+2m(m+n)+(m+n)^2]=(m+n)[2(m+n)^2+(m+n)^2]=3(m+n)^3

3m(m-n)-2n(m-n)的平方

3m(m-n)-2n(m-n)²=(m-n)(3m-2mn+2n²)