C 求Y的值y=X² 1Y=X²-1
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x=3yx+y=12所以x=9,y=3x-y=9-3=6
|x-y|=x-y>0x>y又|x|=1999|y|=2000x=-1999,y=-2000x+y=-3999
2/x+2/y=根号24(2y+2x)/xy=2√6x+y=xy√6x/y(x-y)-y/x(x-y)=1/(x-y)[(x/y-y/x)]=1/(x-y)[(x²-y²)/xy]
你的题目有点问题我这样做了x/{y(x-y)}-y/{x(x-y)}=(x平方-y平方)/{xy(x-y)}=(x+y)/xy2/x+2/y=2(x+y)/xy=根号24
x+y/x-y=1/2取倒数x-y/x+y=2所以x-y/x+y-2x+2y/x-y=x-y/x+y-2(x+y/x-y)=2-2×1/2=2-1=1
x/(y+z)=y/(x+z)=z/(x+y)当x+y+z=0时,x+y=-z(x+y)/z=-z/z=-1当x+y+z≠0时,由x/(y+z)=y/(x+z)=z/(x+y)根据等比性质可得(x+y
(x+y)(x-y)=12因为xy为自然数所以x=4y=2
=-(xy^2)^4+3(xy^2)^3+(xy^2)^2=-27-27-3=-57
x=1,y=1z=x++把x给z,所以z=1,之后x++,所以x=2;y++原来y=1,现在变成2++y前面y=2,现在变成3.所以,最后:x=2y=3,x=1
因为X-Y=1所以原式=x*1+y*(-1)+2013=x-y+2013=1+2013=2014
x(x-y)-y(y-x)=12那么:化简得:(x+y)*(x—y)=12x、y是自然数,所以x1=2,y1=4x2=4,y2=2所以x+y-xy=-2
|x|+x+y=10|y|+x-y=12两式相加得|x|+|y|+2x=22.(1)两式相减得|x|-|y|+2y=-2.(2)所以-|y|+2y<0若y>0,则显然-y+2y<0,即y<0,矛盾若y
即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²
x+y=1x-y=2(x+2y)(x-2y)-(2x-y)(-y-2x)=(x+2y)(x-2y)+(2x-y)(y+2x)=x²-4y²+4x²-y²=5x&
x²+y²=19(x+y)²=x²+y²+2xy=25xy=3(x-y)*(x-y)=(x-y)²=x²+y²-2xy=
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+
7y(x-3y)²-2(3y-x)³=7y(x-3y)²+2(x-3y)³=(x-3y)²[7y+2(x-3y)=(x-3y)²(7y+2X
把x=9y带入到3x=2y+1中得到3*(9y)=2y+127y=2y+125y=1y=1/25x=9y=9/25所以x=9/25,y=1/25
绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=03x-y=0(1)x+y=0(2)(1)+(2)4x=0x=0,代入(2)y=-x=0(x-y)/(xy)无意义,因此题目错了.