已知2x-5y-4=0
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5x-4y=0;x=4y/5;x^2+y^2/x^2-y^2-x+y/x-y=(x²+y²-(x+y)²)/(x²-y²)=(x²+y
括号应该成对,还有一个括号在哪里?能不能把你的叙述用数学的语言表达出来?
很多数学符号,还是写出来方便.
x²+y²-2x+4y+5=0配方:(x-1)²+(y+2)²=0所以只有x-1=0且y+2=0∴x=1,y=-2∴(X^4-y^4)/(2x^2+xy-y
x²+y²-2x+4y+5=0(x²-2x+1)+(y²+4y+4)=0(x-1)²+(y+2)²=0x-1=0x=1y+2=0y=-2x+
∵x²+y²-4x+2y+5=0x²-4x+4+y²+2y+1=0(x-2)²+(y+1)²=0x-2=0y+1=0∴x=2y=-14x
X^2表示平方X^2+4X+4+Y^2-2Y+1=0(X+2)^2+(Y-1)^2=0因为平方大于=0所以X+2=0Y-1=0X=-2Y=1X^2+Y^2=5
(x+2)^2+|x+y+5|=0,x+2=0,x+y+5=0x=-2,y=-33(x^2)y-2(x^2)y+2xy-(x^2)y+4x^2-xy=4x^2+xy=4*(-2)^2+(-2)*(-3
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+4y-2x+8y+5=0,(x-1)+(2y+2)=0∵(x-1)≥0,(2y+2)≥0∴(x-1)=0,(2y+2)=0x=1,y=-1代入x^4-y^4/2x+xy-y·2x-y/xy-y÷(
可能题有误.若x^2+y^2-4x-2y+5=0.(x-2)^2+(y-1)^2=0x=2,y=1根号x+y除以根号下x-y为根3
已知x^2+4y^2-4x+4y+5=0求((y^4-x^4)/(y-2x)(x+y))*((2x-y)/(xy-y^2))/((x^2+y^2)/y)的值答案:x²+4y²-4x
x²+4x+y²-2y+5=0x²+4x+4+y²-2y+1=0(x+2)²+(y-1)²=0所以x+2=0y-1=0则x=-2y=1所以x
(x³+y³)+(x²y+xy²)+(x+y)=0(x+y)(x²-xy+y²)+xy(x+y)+(x+y)=0(x+y)(x²-
2x+y+3=xy,①若x=1,则y不存在,此时5x+4y无最值;②若x≠1,则y=(2x+3)/(x-1),M=5x+4y=5x+(8x+12)/(x-1)=(5x²+3x+12)/(x-
1,-3再问:过程。。。再答:★(x²-2x)+(y²-4y)=5★(x-1)²+(y-2)²=1+4-5★(x-l)²=0,(y-2)²=
x+2=0,x+y+5=0x=-2,y=-33x^2y+{-2X^2y-[-2xy+(x^2y-4x^2)]-xy}=3x^2y-2X^2y+2xy-x^2y+4x^2-xy=4x^2+xy=4*4+
x^2+y^2-2x+4y+5=0(x-1)^2+(y+2)^2=0x=1,y=-2化解式子,再代入
原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009)y=0+(1×1004-2009)y=-1005y=1005/2