y=x2,y=1,x=0围成图形面积详细解答
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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
∵x2+y2-2x-4y+5=0,∴x2-2x+1+y2-4y+4=0,∴(x-1)2+(y-2)2=0,∴x=1,y=2,∴原式=11×2+12×3+…+12011×2012=1-12+12-13+
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
∵2x2-xy-5x+y+4=0∴x2+x2-xy-4x-x+y+4=0∴x2-4x+4+x(x-y)-(x-y)=0∴(x-2)2+(x-y)(x-1)=0∵(x-2)2≥0,x≥y≥1,∴(x-y
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
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
[(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)
∵|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=-
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
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,+∞).
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
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