(ab^2 b^2-3a 1)5 a
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/09 10:21:42
(2-ab+2a+5b)-(3ab+2b-2a)-(3a+4b-ab)=2-ab+2a+5b-3ab-2b+2a-3a-4b+ab=a-b-4ab+2=4-4+2=2
再答:不懂欢迎追问再答:满意麻烦采纳一下谢谢
已知正整数a>b>0满足a²+ab+b²|ab(a+b),求证(a-b)³>3ab.设a,b的最大公约数(a,b)=d,a=md,b=nd.代入条件得(m²+m
ab^2=-2-ab(a^2b^5-ab^3-b)=-a^3b^6+a^2b^4+ab^2=-(ab^2)^3+(ab^2)^2+ab^2=-(-2)^3+(-2)^2+(-2)=8+4-2=10
原式=5a²b(a-b)-3ab(a-b)²-5ab²(a-b)=ab(a-b)[5a-3(a-b)-5b]=ab(a-b)(5a-3a+3b-5b)=ab(a-b)(2
ab^2=-6所以-ab(a^2*b^5-ab^3-b)=-a^3b^6+a^2b^4+ab^2=-(ab^2)^3+(ab^2)^2+(ab^2)=-(-6)^3+(-6)^2+(-6)=216+3
(2a+3b-2ab)-(a+4b+ab)-(3ab-2a+2b)=2a+3b-2ab-a-4b-ab-3ab+2a-2b=3a-3b-5ab=3(a-b)-5ab=3x5-5x(-5)=40
ab(36b-6b-b)=29ab^2=(30-1)*6=174
-ab*(a^2b^5-ab^3-b)=-a^3b^6+a^2b^4+ab^2=-(-2)^3+(-2)^2+(-2)=8+4-2=10
(5a-5b-ab)-(4a-6b-3ab)=5a-5b-ab-4a+6b+3ab=a+b+2ab=3-2x2=-1
(22a+3b-2ab)-(a+4b+ab)-(3ab+2b-2a)=22a+3b-2ab-a-4b-ab-3ab-2b+2a=23a-3b-5ab题目抄差了!(22a+3b-2ab)应该是(2a+3
2a^2b+ab^2-3a^b+5ab-2ab^2-3ab=-a^2b-ab^2+2ab=-ab(a+b)+2ab=-10×1+2×10=10
提取个b你就明白了,
1.3a+2b-5a-b=-2a+b2.3(-ab+2a)-2(3a-ab)=-3ab+6a-6a+2ab=-a
(2a+5b-3ab)-(a+6b-ab)-(2ab+2b-2a)=2a+5b-3ab-a-6b+ab-2ab-2b+2a=3a-3b-4ab=3(a-b)-4ab=12-12=0
-5a^2b+6ab^2-4ab-2a^2b+4ab+3=-5a^2b-2a^2b+6ab^2-4ab+4ab+3=(-5-2)a^2b+6ab^2+3=-7a^2b+6ab^2+36x+2x^2-3
(1)6x+2x-3x+x+1=6x+1其中x=5原式=30+1=31(2)1/3m-3/2n-5/6n-1/6m=1/6m-14/6n=6/1m-7/3n其中m=6,n=3原式=1-7=-6第二题一
ab/a+b=3ab=3(a+b)2ab/a+b-5(a+b)/ab=6(a+b)/(a+b)-5(a+b)/3(a+b)=6-5/3=13/3