Y=x2,y=0,x=1围成的平面图形的面积?并求改曲线绕x轴旋转的体积
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/21 20:02:51
x²-2xy+y²-x+y-1=0(x-y)²-(x-y)-1=0[x-y-(1+√5)/2][x-y-(1-√5)/2]所以x-y=(1+√5)/2或x-y=(1-√5
(x2+y2)/(x-y)=(x2+y2-2xy+2xy)/(x-y)因为xy=1,所以=[(x-y)^2+2]/(x-y)=(x-y)+2/(x-y)因为x>y>0所以(x-y)>0所以有不等式的定
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x−y=0y=x2−2x,解得x=0或x=3,则根据积分的几何意义可知所求图形的面积为:S=∫30(x−x2+2x)dx=∫30(3x−x2)dx=(32x2−13x3)|30=32×32−13×3
x²+(2-y)x+y²-y+1=0方程有解的条件是:△=b²-4ac≥0→-3y²≥0∴y=0∴x=-1
因为y=3x/(x²+x+1)所以1/y=(1/3)x+(1/3)+(1/3)/x因为x
[(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)
x²+y²-10x-6y+34=0,即(x-5)²+(y-3)²=0,所以,x=5,y=3,所以(x+y)/y=(5+3)/3=8/3
x^2+(2-y)x+y^2-y+1=0这个关于x的二次方程有解b^2-4ac>0-3y^2>0所以y=0x=-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
由题意阴影部分的面积为∫120x2dx+∫11214dx=13x3| 120+14x|112=124+14−18=16;故选A.
由x2-1=0,得抛物线与轴的交点坐标是(-1,0)和(1,0),所求图形分成两块,分别用定积分表示面积S1=∫1−1|x2−1|dx,S2=∫21(x2−1)dx.故面积S=S1+S2=∫1−1|x
用均值不等式,考虑X>0,X
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
因为(x+y)(x+y+1)=0所以x+y=0,或x+y=-1,x2+y2+2xy=(x+y)²当x+y=0时,原式=0,当x+y=-1时,原式=1
∵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)