将y=1 (x^2 3x 2)展成x的幂级数,并指出收敛区间
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![将y=1 (x^2 3x 2)展成x的幂级数,并指出收敛区间](/uploads/image/f/3951636-60-6.jpg?t=%E5%B0%86y%3D1+%28x%5E2+3x+2%29%E5%B1%95%E6%88%90x%E7%9A%84%E5%B9%82%E7%BA%A7%E6%95%B0%2C%E5%B9%B6%E6%8C%87%E5%87%BA%E6%94%B6%E6%95%9B%E5%8C%BA%E9%97%B4)
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
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
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
将原方程变形为x2+x+1x2+1+x2+1x2+x+1=23+32.设y=x2+x+1x2+1,则原方程变为y+1y=23+32,解得y1=23,y2=32.当x2+x+1x2+1=23时,x=-3
原抛物线的顶点坐标为(0,0),新抛物线的顶点坐标为(1,-1),∴将原抛物线向右平移1个单位,再向下平移1个单位可得到新抛物线.故选D.
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>=
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^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)
因式分解x²(x+1)-y(xy+x)原式=x³+x²-xy²-xy=x³-xy²+x²-xy=x(x²-y²
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
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
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)