y=x-3 2x-x2
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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
#includemain(){intx,y;charch='*';printf("输入x的值:");scanf("%d",&x);if(x>0){y=x+1;}elseif(x
y=a(x-x1)(x-x2)与x轴的交点是:(x1,0);(x2,0)y=a(x+x1)(x+x2)与x轴的交点是:(-x1,0);(-x2,0)再问:他们一样吗?再答:不一样。交点的横坐标分别是相
X²(X+1)-Y(XY+X)=X^3+X²-XY²-XY=X^3-XY²+X²-XY=X(X²-Y²)+X(X-Y)=X(X-Y
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+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
因为y=3x/(x²+x+1)所以1/y=(1/3)x+(1/3)+(1/3)/x因为x
1、y=(x²-3x)/(x+1)那么y'=[(x²-3x)'*(x+1)-(x²-3x)*(x+1)']/(x+1)²显然(x²-3x)'=2x-3
就把(x2+1/x)=y代进去(2(x2+1)/x)+(6x/x2+1)=7化为2y+6/y=7解之得y=3/2或2(x2+1/x)=3/2时无解(x2+1/x)=2时x=1再问:用换元法把(x2+1
对x求导0.5*1/(x²+y²)*(x²+y²)'=1/[1+(y/x)²]*(y/x)'0.5/(x²+y²)*(2x+2y*
(X+Y)²=X²+Y²+2XY=X²+Y²+X²-Y²=2X²(X-Y)²=X²+Y²-
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,+∞).
用均值不等式,考虑X>0,X
要使函数f(x)有意义,则x2−2x>0x2−3x+2≥0,即x>2或x<0x≥2或x≤1,解得x>2或x<0,故函数的定义域为(2,+∞)∪(-∞,0)故答案为:(2,+∞)∪(-∞,0)
不对=x2(x-y)-y2(x-y)=(x2-y2)(x-y)=(x+y)(x-y)2再问:噢。我看懂了
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
∵y=1/(x²-x)∴x²-x≠0x(x-1)≠0∴x≠0或x≠1∴定义域为:(负无穷,0)∪(0,1)∪(1,正无穷)
(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