若2a b=12,其中a≥0,b≥0,P=3a 2b.试确定P的最小值和最大值.
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(a-1)²+|b+1|=0a-1=0、b+1=0解得:a=1、b=-13ab-2(a²-ab)-(a²-ab)=3ab-3(a²-ab)=3ab-3a
(b)/(a-b)-(b^3)/(a^3-2a^2b+ab^2)÷(ab+b^2)/(a^2-b^2)=(b)/(a-b)-(b^3)/[a(a-b)^2]*(a+b)(a-b)/b(a+b)=b/(
(a/(ab-b^2)-b/(a^2-ab))/((1+a^2+b^2)/2ab)=(a/(b*(a-b))-b/(a*(a-b)))/((1+a^2+b^2)/2ab)=((a^2-b^2)/(a*
=a^2b+3ab^2-a^2b-4ab^2+2a^2b=2a^2b-ab^2=ab(2a-b)因为|a-1|+(b+2)^2=0所以a=1b=-2=-2(2+2)=-8再问:为什么a=1,b=-2再
a*a*b*2ab+5a*ab#4ab=(a+a+b+2ab)+(5a+ab-4ab)=7a+b-ab=7*5+3-5*3=35+3-15=23
∵a2+ab-2b2=0∴(a+2b)(a-b)=0∴a=-2b或者a=b原式=(a²-b²-a²-b²)/ab=-2b²/ab=-2b/a当a=-2
[a/(a+b)-b/(b-a)-2ab/(a^2-b^2)]/(1/a-1/b)=[a(a-b)+b(a+b)-2ab]/(a^2-b^2)/[(b-a)/(ab)]=[a^2-ab+ab+b^2-
3a^2+ab-2b^2=0(3a-2b)(a+b)=0可得:3a=2b,则2b=3a,a+b=0,则b=-a,所以a/b-b/a-(a^2+b^2)/ab=a/b-b/a-a/b-b/a=-2b/a
(a+b)的平方+(2b-1)的绝对值=0∴a+b=02b-1=0∴a=-1/2b=1/2ab-[2ab-3(ab-1)]=ab-2ab+3ab-3=2ab-3=-1/2-3=-7/2
(a+2)²+|b-1|=0则a+2=0,b-1=0则a=-2,b=1-(3a²-4ab)+[a²-2(2a+2ab)]=-3a²+4ab+a²-4a
-ab*(a^2b^5-ab^3-b)=-a^3b^6+a^2b^4+ab^2=-(-2)^3+(-2)^2+(-2)=8+4-2=10
从今年将拿出就个人个人很听话再问:无聊再答:哎刚刚做个任务而已啦。。。原式=[a(9a-6b)+b²]/[a(3a-b)]把a约掉=[9a-6b+(b²/a)]/(3a-b)=a[
(a/ab-b^2-b/a^2-ab)/(1+a^2+b^2/2ab),其中a=-2,b=1=[a/b(a-b)-b/a(a-b)]/[(2ab+a²+b²)/2ab]=[(a
a*3-b*3=3a*2b-3ab*2+1(a-b)(a^2+b^2+ab)-3ab(a-b)=1(a-b)(a^2+b^2+ab-3ab)=1(a-b)(a-b)^2=1(a-b)^3=1a-b=1
(1/ab)(a+b-4/(a+b))ab=15a-b=2(a-b)^2=4(a+b)^2=(a-b)^2+4ab=64a+b=8(ab=15>0,a-b=2>0所以a+b>0)(1/ab)(a+b-
先化简,再求值;[(2a-b/a+b)-(b/a-b)]÷{(a-2b/a+b)+[ab/(a-b)²]},其中b/a=-1/2解,得:[(2a-b)(a-b)-b(a+b)]/[(a+b)
∵ab=-3.∴b=-3/a.将b=-3/a代入a+2b=-5得:a+2(-3/a)=-5a-6/a=-5a-6=-5aa+5a-6=0(a+6)(a-1)=0∵a+6=0∴a1=-6;∵a-1=0∴
(3ab-2b)+[3a-(5ab-12b-2b)],﹦3ab-2b+3a-5ab+14b=-2ab+3﹙a+2b﹚+6b其中a+2b=-5,ab=-3b﹙-5-2b﹚=﹣3b=1/2,或b=-3所以
原式=-(9a^4-18a^3+3b^2+4a^3b)=-(9a^4-14a^3b+3b^2)=-(9*1-14*(-1)^3*(-2)+3*(-2)^2)=-(9-28+12)=7
原式=(2a-2b+3ab-3b)+(a-b+2a)-(8a-8b+ab-b)=8+9-3b+4+2a-32+3+b=-8+2a-2b=0还可以解出a和b的值带入