x=2y-1 3x y=11

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x=2y-1 3x y=11
已知x-xy=8,xy-y=-9,求x+y-2xy的值

x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17

解方程组x+y+xy=7 x^2+y^2+xy=13

第一题设x+y=axy=b则x^2+y^2=a^2-2b原方程变为a^2-2b+b=13①b+a=7②这就变成了简单的二元二次方程由②得b=7-a带入①得a^2+a-20=0(a-4)(a+5)=0得

如果x^2+xy+y-14,y^@+xy+x=28,求x+y的值.

x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7

(-3x^y+2xy)-( )=4x^+xy

(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^

x^2+y^2=7,xy=-2,求5x^2-3xy-4y^2-11xy-7x^2+2y^2的值

5x²-3xy-4y²-11xy-7x²+2y²=-2x²-2y²-14xy=-2(x²+y²)-14xy=-2×7-1

已知x+y分之xy=2,那么3x-5xy+3y分之3xy-x-y=

xy/(x+y)=2∴xy=2(x+y)(3xy-x-y)/(3x-5xy+3y)=[6(x+y)-x-y]/[3x+3y-10(x+y)]=5(x+y)/[-7(x+y)]=-5/7

x-xy=40,xy-y= -20,求代数式x-y和x+y-2xy

(x-xy)+(xy-y)=40-20x-xy+xy-y=20x-y=20(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=60x+y-2xy=60

已知:xy+x=-1,xy-y=-2.

(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

x^2+xy+x=36,y^2+xy+y=20,求x+y.

7或者-8再问:求过程^_^再答:两个等式两边相加

已知x^2+y^2=7,xy=-1,求代数式:5x^2-3xy-4y^2-11xy+2y^2-7x^2

原式=-2x^2-2y^2-3xy-11xy=-2*7-14*(-1)=0

由已知x+y=-2,xy=3那么2(x+xy)-[(xy-3y)-x]-(-xy)等于多少?

2(x+xy)-[(xy-3y)-x]-(-xy)=2x+2xy-xy+3y+x+xy=3x+3y+2xy=3(x+y)+2xy=3*(-2)+2*3=0

(x+2y-3xy)-(-2x-y+xy),其中x+y=1/2,xy=-1/2

(x+2y-3xy)-(-2x-y+xy)=x+2y-3xy+2x+y-xy=(1+2)x+(2+1)y-(3+1)xy=3x+3y-4xy=3(x+y)-4xy=3*1/2-4*(-1/2)=3/2

已知x-y=4xy,则2x+3xy-2yx-2xy-y

∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.

X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=

X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=(x+y)²/xy×(x-y)²/y(x+y)=(x+y)(x-y)&#

x²+xy=12 xy+y²=13

两式相加得到x+y=5,相减得y-x=1/5,故x=12/5,y=13/5xy=156/25,因为要求的都是正数,而且xy同正负,所以只考虑x,y正数即可故x²+y²=(x+y)^

已知x^2+y^2=7,xy=-1,求代数式:5x^2-(3xy+4y^2)-(11xy-2y^2)-7x^2

5x^2-(3xy+4y^2)-(11xy-2y^2)-7x^2=5x^2-3xy-4y^2-11xy+2y^2-7x^2=-2x^2-2y^2-14xy=-2(x^2+y^2)-14*xy=-2*7

(x+2y-3xy)-(-2x-y+xy),其中x+y=,xy=-,求值.

3x-4xy+3y再问:然后的过程再答:再带入值等于3.5再答:马上再答:给你照片再答:再答:亲采纳哟

已知x-2y=3,x²-2xy+4y²=11,求①xy ②x²y-2xy²

∵x-2y=3∴(x-2y)²=9即x²-4xy+4y²=9∵x²-2xy+4y²=11两式相减:2xy=2∴xy=1又x²y-2xy

已知x^2-xy=14,xy-y^2=-11,求x^2-2xy+y^2的值

x^2-xy=14,(1)xy-y^2=-11,(2)(1)-(2)得:x^2-2xy+y^2=14-(-11)=25