16(x 2)²-49(x-y)²

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函数y=(x2-x)/(x2-x+1)的值域

去分母得:x^2(y-1)+x(1-y)+y=0y=1时,上式无解y=1时,为二次式,须有delta>=0即(1-y)^2-4y(y-1)>=0(y-1)(3y+1)再问:x^2(y-1)+x(1-y

求教 y={ x+1(x>0) x2 (x

#includemain(){intx,y;charch='*';printf("输入x的值:");scanf("%d",&x);if(x>0){y=x+1;}elseif(x

XY(XY-X2)乘以------------X-Y

XY(XY-X^2)(X-Y)=X^2Y(X-Y)(X-Y)=(X-Y)(X^2Y-1)再问:答案是。-X2Y.我不知道过程是如何来的。能告诉我过程吗。谢谢。负X的平方Y

因式分解x2+y-xy-x

x²+y-xy-x=﹙x²-xy﹚+﹙y-x﹚=x﹙x-y﹚-﹙x-y﹚=﹙x-y﹚﹙x-1﹚

当x,y≥0时,曲线x2+y2=|x|+|y|互为x2

当x,y≥0时,曲线x2+y2=|x|+|y|互为x2+y2=x+y,曲线表示以(12,12)为圆心,以22为半径的圆,在第一象限的部分;当x≥0,y≤0时,曲线x2+y2=|x|+|y|互为x2+y

已知|x-y+1|与x2+8x+16互为相反数,求x2+2xy+y2的值.

∵|x-y+1|与x2+8x+16互为相反数,∴|x-y+1|与(x+4)2互为相反数,即|x-y+1|+(x+4)2=0,∴x-y+1=0,x+4=0,解得x=-4,y=-3.当x=-4,y=-3时

x2 (x+y+z+xyz)因式分解

因式分解分为以下四种情况:提取公因式法,乘法公式法,分组分解法,十字相乘法.此题不符合任一形式,所以不能再分解.

已知x、y满足x2+y2+174

将x2+y2+174=4x+y,变形得:(x2-4x+4)+(y2-y+14)=0,即(x-2)2+(y-12)2=0,解得:x=2,y=12,则原式=2×122+12=25.

已知x-y=5,(x+y)2=49,则x2+y2的值等于______.

∵x-y=5,∴x2+y2-2xy=25①,∵(x+y)2=49,∴x2+y2+2xy=49②,∴①+②得:2(x2+y2)=74,∴x2+y2=37.故答案为:37.

若x2+4y2+2x-4y+2=0求5x2+16y2的算术平方根

x²+4y²+2x-4y+2=0(x²+2x+1)+(4y²-4y+1)=0(x+1)²+(2y-1)²=0x=-1,y=1/25x

已知x-y=5,(x+y)2-49,求X2+y2的值

∵x-y=5∴x²-2xy+y²=25∵(x+y)²=49∴x²+2xy+y²=49两式相加可得2x²+2y²=74∴x²

因式分解 x2(x+1)-y(xy+x)

因式分解x²(x+1)-y(xy+x)原式=x³+x²-xy²-xy=x³-xy²+x²-xy=x(x²-y²

分解因式:(1)x2(x-y)+y2(y-x);(2)(x+y)2+64-16(x+y).

(1)x2(x-y)+y2(y-x)=(x-y)(x2-y2)=(x-y)2(x+y);(2)(x+y)2+64-16(x+y)=(x+y-8)2.

函数y=4x2+1x

解析:y′=8x-1x2=8x3−1x2,令y′>0,解得x>12,则函数的单调递增区间为(12,+∞).故答案:(12,+∞).

y=2X(X2+16)(X>0)的最值.(X2即X的二次方)

没有最大值,也没有最小值.因为,x+16在x>0上是增函数;2x在x>0上也是增函数.所以,该函数既没有最大值,也没有最小值.

我做对了吗.x2(x-y)+y2(y-x)= x2(x-y)-y2(x-y)=x2-y2=(x+y)(x-y)

不对=x2(x-y)-y2(x-y)=(x2-y2)(x-y)=(x+y)(x-y)2再问:噢。我看懂了

X×Y=6,X2+Y2=16,则X=?,Y=?,或者X+Y=?

X2+Y2=16;X2+Y2+2XY=(X+Y)2;因为X2+Y2=16,XY=6所以16+12=(X+Y)2;28=(X+Y)2;X+Y=根号28

x2-x-y2-y 解法:=(x2-y2)-(x+y) =(x+y)(x-y)-(x+y) =(x+y)(x-y-1)

哥!你那个是x方y方吧!有这么个公式x方-y方=(x+y)(x-y)所以得到了(x+y)(x-y)-(x+y)这时候提取公因式(x+y)就得到了(x+y)(x-y-1)再问:是啊,怎么提(X+Y)他那