已知:x-y≤1,x y≤1, 2x-y≥1,求:3x-2y最小值
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![已知:x-y≤1,x y≤1, 2x-y≥1,求:3x-2y最小值](/uploads/image/f/4207336-16-6.jpg?t=%E5%B7%B2%E7%9F%A5%3Ax-y%E2%89%A41%2Cx+y%E2%89%A41%2C+2x-y%E2%89%A51%2C%E6%B1%82%3A3x-2y%E6%9C%80%E5%B0%8F%E5%80%BC)
=-x-(2y-2+3x)+2(x+4)=-x-2y+2-3x+2x+8=-4x-2y+10
x^2+1/2y⒉+4≤xy+2yx^2+1/2y^2+4-xy-2y
1.x^4+5x^2-6=(x^2+6)(x^2-1)=(x^2+6)(x-1)(x+1)2.(x+y)(x+y+2xy)+(xy+1)(xy-1)=(x+y)^2+2xy(x+y)+(xy+1)(x
(1)已知x+y=6,xy=-3x^2y+xy^2=xy(x+y)=-3*6=-18(2)x*[2-(1/x)]+(x/x^2-2x)*(x^2-4)=x*(2x-1)/x+[x/x(x-2)]*(x
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-6xy+3x-3y=-6×(-2)+3×1=15
答:x+y=-1,xy=-2-5(x+y)+(x-y)+x(xy+y)=-5x-5y+x-y+xy(x+1)=-4x-6y+(-2)(x+1)=-4x-6y-2x-2=-6x-6y-2=-6(x+y)
x^2y+xy^2=xy(x+y)=1/5
(1)∵xy+x=-1①,xy-y=-2②,∴①-②得x+y=1;(2)先把xy+x=-1,xy-y=-2的值代入代数式,得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8
画出限定区域z=2x-yy=2x-z当y=2x-z经过点C时,z有最大值z最大值=2*2-(-1)=5如果您认可我的回答,请点击“采纳为满意答案”,祝学习进步!再问:�Ͽ��Ͽɣ���Щ֪ʶ�Ǹ���
3/(x-y)=1/xyx-y=3xyy-z=-3xy原式=[(y-x)-2xy]/[2(x-y)+3xy]=[(-3xy)-2xy]/[2(3xy)+3xy]=-5xy/9xy=-5/9
-5(x+y)+(x-y)+2(xy+y)=-5x-5y+x-y+2xy+2y=-4x-4y+2xy=-4(x+y)+2xy=-4×(-1)+2×(-2)=4+(-4)=0你有问题也可以在这里向我提问
先把y≤xx+y≤1y≥-1画出来具体是画y=x,y=-x+1,y=-1三条直线而范围是y≤x即y=x的下面(等号包括直线本身,)同理x+y≤1,y≤-x+1即y=-x+1的下面如图再看z=2x-y变
x²+y²-xy+2x-y+1=[3(x+1)²+(x-2y+1)²]/4=0,由于(x+1)²>=0且(x-2y+1)²>=0,则有x+1
这道题目还是在考察韦达定理的运用用伟大定理求出xy的值再代入代数式否则是求不出来的(x+y)^2=x^2+y^2+2xy=1x^2+y^2=5(x-y)^2=5-2(-2)=9下面分两种情况讨论1x-
因为xy/(x+y)=1/2所以x+y=2xy原式=3(x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1
由于这是选择题,可以进行分析而不一定求解.首先,①式+②式得:2x≥2,即x≥1.④将y=(3/2)x代入①式,得x≤-2,与④式不符,所以(A)不是正确答案;将y=(5/2)x代入①式,得x≤-2/
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=(-2xy-3xy-xy)+(2x+2x-x)+(3y-2y-4y)=-6xy+3x-3y=-6+3*3=3
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6*(-2)+3*1=15不懂可追问,有帮助请采