若x²+y²=12,x y=4那么(x-y)²=?
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|x-1|+y^2+4y+4=|x-1|+(y+2)^2=0因为|x-1|>=0,(y+2)^2>=0则:x-1=0y+2=0x=1y=-2xy=-2
5x²-3xy-4y²-11xy-7x²+2y²=5x²-7x²+2y²-4y²-3xy-11xy=-2x²-
(3xy)平方/(-xy)+(x-2y)平方-(x+2y)(x-2y)=-9xy+(x-2y)(x-2y-x-2y)=-9xy+(x-2y)(-4y)=-9xy-4xy+8y²=8y
2x+y+6≥6+2√2xyxy≥6+2√2xy(√xy-√2)^2≥8√xy-√2≥2√2或√xy-√2≤-2√2(不可能)所以xy最小值是(3√2)^2=18-------------------
即y=-2x所以原式=(x²-2x²+4x²)/(-4x²-x²)=3x²/(-5x²)=-3/5
由题设可知y=xy-1,∴x=yx3y=x4y-1,∴4y-1=1,故y=12,∴12x=x,解得x=4,于是x+y=4+12=92.故答案为:92.
x(x+y)(x-y)-x(x+y)2=x(x+y)(x-y-x-y),=x(x+y)(-2y),=-2xy(x+y),当x+y=1,xy=-12时,原式=-2×(-12)×1=1.
像这种题目也可以用特殊值去做,比如令x=1,代入方程求出y,再代入所求的式子就可以得出答案.
(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6xy+3*(x-y)当时,原式=-6*1+3*
4x²+8xy+9y²=2(2x²+xy)+3(3y²+2xy)=38.
解题思路::∵x+y=0,x+13y=1,解得x=1/12,y=-1/12∴x²+12xy+13y²=1/144-1/12+13/144=14/144-1/12=2/144=1/72解题过程:已知x+
求x的平方y+xy的2次方=xy(x+y)=30
xy≦(x^2+y^2)/2,当x=y时等号成立,这时xy取最大值;因为x+y=4,所以当x=y时,x=y=2,所以xy的最大值为(x^2+y^2)/2=4.再问:其他方法呢?再答:(1)x+y=4,
A=5x^2y-3xy^2+4xyB=7xy^2-2xy+x^2y2C=-A-B2C-2A=-3A-B=-15x^2y+9xy^2-12xy-7xy^2+2xy-x^2y=-16x^2y+2xy^2-
(3x-2y+8xy)-(x-4y-5xy)=3x-2y+8xy-x+4y+5xy=(3x-x)+(-2y+4y)+(8xy+5xy)=2x+2y+13xy=2(x+y)+13xy=2×3+13×(-
再答:请好评、保对
x^2·y+2x^2·y^2+xy^2=xy×(x+2xy+y)=3×(5+6)=33谢谢~
x^2+y^2+4x+6y+13=0x^2+y^2+4x+6y+4+9=0x^2+4x+4+y^2+6y+9=0(x+2)^2+(y+3)^2=0(∵(x+2)^2>=0,(y+3)^2>=0)=>(
∵有理数x,y满足xy≠0,∴x|x|=±1,|y|y=±1,∴m=x|x|+|y|y的最大值是m=1+1=2.故答案为:2.