如何画y=| x-1| |x 2|图像

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如何画y=| x-1| |x 2|图像
x(x-1)-(x2-y)=-3,求x2-y2-2xy的值

解题思路:由完全平方公式可求解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/rea

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

已知(x-x2)+(x2-y)=1,求代数式12(x

∵(x-x2)+(x2-y)=1,∴x-y=1.∵12(x2+y2)-xy=12(x2+y2-2xy+2xy)-xy=12(x2+y2-2xy)+xy-xy=12(x-y)2,把x-y=1代入上式,得

因式分解X2(X+1)-Y(XY+X)=

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

求函数Y=(x2-x+1)/(x2+x+1)值域

yx²+yx+y=x²-x+1(y-1)x²+(y+1)²+(y-1)=0x是实数则方程有解所以△>=0所以(y+1)²-4(y-1)²>=

求y=3x/x2+x+1的值域.

因为y=3x/(x²+x+1)所以1/y=(1/3)x+(1/3)+(1/3)/x因为x

求函数y=x2+2x+1/(x2+2x+3)的最小值

y=x2+2x+1/(x2+2x+3)=(x+1)2/(x2+2x+3)当分母一定时,分子越小越好(x2+2x+3)=(x+1)2+2永远大于零当(x+1)2越小越好而X=-1时y=x2+2x+1/(

函数y=x2-3x/x+1求导

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

已知[(x2+y2)-(x-y)2+2y(x-y)]÷4y=1,求4x/4x2-4y2-1/2x+y的值,

[(x^2+y^2)-(x-y)^2+2y(x-y)]÷4y=1(x^2+y^2-x^2+2xy-y^2+2xy-2y^2)÷4y=1(4xy-2y^2)4y=12x-y=24x/(4x^2-y^2)

y=|x2-2x|+1 画出此函数的图像 x2是x的平方

先画出y=x^2-2x的图像,然后把在x轴下方翻折到上面,然后总体向上平移一个单位由于不方便所以只能告诉你方法再问:y=x^2-2x用不用考虑两种情况?再答:不用,那样会很麻烦

用换元法解(2(x2+1)/x)+(6x/x2+1)=7其中(x2+1/x)=y

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

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,+∞).

函数y=x2-1/x2+1的值域(x2为x的平方)

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

求函数y=x/x2+x+1的值域

用均值不等式,考虑X>0,X

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

2x(x-y)4-x2(x-y)2+xy(y-x)2 如何因式分解

原式=2x(x-y)4-x2(x-y)2+xy(x-y)2=x(x-y)2{2(x-y)2-x+y}=x(x-y)2{2(x-y)2-(x-y)}=x(x-y)3(2x-2y-1)

求函数y=x-1/x2-x的定义域

∵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)