y=-0.5x²-1.5x 2化为顶点式
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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
由N圆方程设x=3cost,y=a+3sint,(0=
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
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
∵|x-y+1|+(x-5)2=0,则x-y+1=0,x-5=0,解得x=5,y=6.(-3x2-4y)-(2x2-5y+6)+(x2-5y-1)=-3x2-4y-2x2+5y-6+x2-5y-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*
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,+∞).
还是按照你的记法:x2为x的平方,y=(x2-1)/(x2+1)两边同乘以x2+1得:y(x2+1)=x2-1去括号y*x2+y=x2-1移项y*x2-x2+y+1=0(y-1)x2+y+1=0x为实
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)
解题思路:公切线方程解题过程:
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
-8x2-5x(-x+3y)+(3x+2y)(x-y),其中x=2,y=-1.=-8x²+5x²-15xy+3x²-xy-2y²=-16xy-2y²=
∵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)