x2 y=x^2 4 问 等于多少时使得当
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5x2y+(-6x2y)+34x2y=14x2y答:和是-14x2y.
(x-8)+(x-9)=x+3x-8+x-9-x-3=0x=8+9+3x=17+3x=20
原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.
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^2=xx=0.25=0.5X0.5x=0.5
X+X+X+X+X+X+100X=1690106x=1690x=845/53
化简得:9-12Y^2+6Y+4+12Y^2+4Y-10-10Y+X-Y+1=X-Y+4带入X、Y值得:=3
x=2013,y=2013[3x(x²y-xy²)+xy×(3xy-2x²)]/(x²y)=[3x²y(x-y)+x²y×(3y-2x)]/
原式=x4+x3y+4x3y+x2y+4x2y2+4x2y2+xy2+4xy3+xy3+y4,=x3(x+y)+4x2y(x+y)+xy(x+y)+4xy2(x+y)+y3(x+y),=-x3-4x2
取δ=0.0002,当0
2x=100-20=80x=80/2=40回答完毕
x^2-y^2/xy=x/y-y/x=2-1/2=1.5
这题目xyz难道没有约束条件?如果x,y,z都是正整数的话,由于231正约数为3,7,11所以x+y+z=3+7+11=21如果x,y,y只是整数,就需要考虑正负问题.可以为-3+7-11=-7,-3
x=±1,y=±3,z=±2xyzz>y则0>x>z>yx=-1,y=-3,z=-2,x2y-[4x2y-(xyz-x2z)-3x2z]-2xyx=x2y-4x2y+xyz-x2z+3x2z-2xyx
X=60~~~~~~~~~
代入x=-1,y=1,2x^y-(5xy^-3x^y)-x^=2*(-1)^*1-{5*(-1)*1^-3*(-1)^*1}-(-1)^=2-(-5-3)-1=9备注:2^表示2的平方
f(x-1)=x²+4x-5=(x-1)²+6(x-1),f(x+1)=(x+1)²+6(x+1)=x²+8x+7.方法类似.直接看不出来的题目可以换元,比如这
解3xy²-[2xy²-2(xy-1.5x²y)]+xy-3x²y=3xy²-(2xy²-2xy+3x²y)+xy-3x²