已知x-y 3-x y 2=1, 用含x的代数式表示y 用含y的代数式表示x
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x3+y3=100(x+y)(x^2-xy+y^2)=100因x+y=1所以x^2-xy+y^2=100(x+y)^2-3xy=1001-3xy=100xy=-33x^2+y^2=(x+y)^2-2x
因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10
∵x+y=1,∴x3+y3+3xy=(x+y)(x2-xy+y2)+3xy=x2+y2+2xy=(x+y)2=1.
解题思路:本题第一个方程容易列出,关键是由第二个条件,经过化简列出第二个方程。解题过程:解:由题意,得x+y=14①∵x3+x2y-xy2-y3=0∴x2(x+y)-y2(x+y)=0∴(x+y)2(
2(xy-5xy2)-(3xy2-xy)=(2xy-10xy2)-(3xy2-xy)=2xy-10xy2-3xy2+xy=(2xy+xy)+(-3xy2-10xy2)=3xy-13xy2,∵(x+1)
(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y-y3=-2y3=-2×(-1)3=2.因为化简的
A+B+C=(x3+3x2y-5xy2+6y3-1)+(y3+2xy2+x2y-2x3+2)+(x3-4x2y+3xy2-7y3+1)=(1+1-2)x3+(3+1-4)x2y+(-5+2+3)xy2
原式=x3+3x2y-5xy2+6x3+1-2x3+y3+2xy2+x2y+2-4x2y-7x3-y3+4xy2+1=-2x3+xy2+4,由于y为偶次幂,故误把“x=3,y=-1”写成“x=3,y=
1=x+2y=x+y+y≥3(xy²)^(1/3).===>xy²≤1/27.等号仅当x=y=1/3时取得,∴(xy²)max=1/27
解题思路:本题第一个方程容易列出,关键是由第二个条件,经过化简列出第二个方程。解题过程:
原式=5xy2-2x2y+3xy2-2x2y=8xy2-4x2y,∵(x-2)2+|y+1|=0,∴x-2=0,y+1=0,即x=2,y=-1,则原式=16+16=32.
(x+2)²+|y-1|=0平方数与绝对值都是非负数两个非负数的和为0,那么这两个数都是0x+2=0y-1=0解得:x=-2,y=1x³+3x²y+3xy²+y
3xy2(x-x3y2-12x2y)=3x2y2-3x4y4-32x3y3,当xy=-1时,原式=3×(-1)2-3×(-1)4-32×(-1)3=32.
A-B=(x3+2y3-xy2)-(﹣y3+x3+2xy2)=x³+2y³-xy²+y³-x³-2xy²=3y³-3xy²
(x+2)²+|y-1|=0平方数与绝对值都是非负数两个非负数的和为0,那么这两个数都是0x+2=0y-1=0解得:x=-2,y=1x³+3x²y+3xy²+y
x3-y3-x2y+xy2=(x-y)(x2+xy+y2)-xy(x-y)=(x-y)(x2+xy+y2-xy)=(x-y)(x2+y2)
x3+3xy-y3=(x-y)(x^2+y^2+xy)+3xy=-x^2-y^2+2xy=-(x-y)^2=-1
x²-x=7y²-y=7相减x²-x-y²+y=0(x+y)(x-y)=x-yx-y≠0约分x+y=1x²-x=7y²-y=7相加x&sup