若x2+2xy+y2-(x+y)·a+25是一个完全平方式,则a的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 06:15:54
![若x2+2xy+y2-(x+y)·a+25是一个完全平方式,则a的值](/uploads/image/f/6966781-61-1.jpg?t=%E8%8B%A5x2%2B2xy%2By2-%28x%2By%29%C2%B7a%2B25%E6%98%AF%E4%B8%80%E4%B8%AA%E5%AE%8C%E5%85%A8%E5%B9%B3%E6%96%B9%E5%BC%8F%2C%E5%88%99a%E7%9A%84%E5%80%BC)
x²-2xy+y²-x+y-1=0(x-y)²-(x-y)-1=0[x-y-(1+√5)/2][x-y-(1-√5)/2]所以x-y=(1+√5)/2或x-y=(1-√5
x-y=1,(x-y)^2=x^2+y^2-2xy=12xy=25-1=24,xy=12(x+y)^2=x^2+y^2+2xy=25+24=49x^2-xy+y^2=25-12=13
首先,X^2-XY+2Y^2=(X-2Y)(X+Y)所以,设x^2-xy-2y^2-x+5y-2可分解成(X-2Y+A)(X+Y+B)则展开有X的一次项=A+B=-1Y的一次项有A-2B=5连列成方程
25(x²+2xy+y²)-9(x²-2xy+y²)=[5(x+y)]²-[3(x-y)]²=[5(x+y)+3(x-y)][5(x+y)-
设:S=x-2y,x=S+2y代入x²+y²=1中得:(s+2y)²+y²=15y²+4sy+s²-1=0∵y是实数,∴△≥0(4s)&su
∵x<y<0,∴x-y<0,x+y<0.∴x2−2xy+y2=(x−y)2=|x-y|=y-x.x2+2xy+y2=(x+y)2=|x+y|=-x-y.∴x2−2xy+y2+x2+2xy+y2=-2x
已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*
x^2+xy-2y^2-x+7y-6=(x+2y)(x-y)-x+7y-6x+2y-3Xx-y2十字相乘得:2(x+2y)-3(x-y)=-x+7y是一次项所以原式=(x+2y-3)(x-y+2)
x2-2xy+y2+3x-3y+2=(x-y)2+3(x-y)+2=(x-y-1)(x-y-2).
10拆成1+9X2-2X+1+Y2-6Y+9=0(X-1)2+(Y-3)2=0平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.所以两个都等于0所以X-1=0,Y-3=0X=1,Y=
x²-2x+y²+6y+10=0,变换得(x-1)²+(y+3)²=0,∴x=1,y=-3∴(x2-2xy)/(xy+y2)=(1²-2*(-3))/
(X+Y)²=X²+Y²+2XY=X²+Y²+X²-Y²=2X²(X-Y)²=X²+Y²-
一原式可化为(x+1)²+(y-3)²=0所以x=-1,y=3二原式可化为(a²-b²)²+(b²-c²)²+(c&su
x²-2xy+y²/x²-y²=(x-y)²/(x-y)(x+y)=(x-y)/(x+y)因为x=3,y=-5,所以(3-(-5))/(3+(-5))
x^2-y^2=2xy,得x/y-y/x=2,即(y/x)^2+2(y/x)-1=0∴y/x=-1+√2或y/x=-1-√2(舍去,因为x,y都是正数).即(x-y)/(x+y)=√2-1
2x²+xy-3y²+x+4y-1=2×x×x+x×y-3×y×y+x+4×y-1=x×2x+x×y-y×3y+x×1+y×4-1=x×(2x+y+1)-y×(3y+4)-1再问:
首先,X^2-XY+2Y^2=(X-2Y)(X+Y)所以,设x^2-xy-2y^2-x+5y-2可分解成(X-2Y+A)(X+Y+B)则展开有X的一次项=A+B=-1Y的一次项有A-2B=5连列成方程
因为x2+xy-2y2=2(x+2y)即(x-y)(x+2y)=2(x+2y)所以x-y=2(x-y)2=x2-2xy+y2=4所以xy=(x2+y2-4)/2=(20-4)/2=8(x+y)2=x2
(x-y)/(x+y)=(x-y)(x+y)/[(x+y)^2]=(x^2-y^2)/[x^2+y^2+2xy]=2xy/[x^2+y^2+x^2-y^2]=2xy/(2x^2)=y/xx^2-y^2
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²