x2y (x3 y3)极限

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已知点a(x1y1),B(x2y2)c(x3y3)都在反比例函数y=-4/x的图像上

∵反比例函数y=-4/x的图像在第2、4象限,∴当x1<0时,y1>0当 0<x2<x3时,图象在第四象限,∴y随x的增大而增大,且y<0∴0〉y3>y2综合起来,有y2<y3<0<y1

化简求值①化简:4ab+8-2b2-9ab-6②先化简,再求值:已知3x2y-[2x2y-3(2xy-x2y)-xy],

(1)4ab+8-2b2-9ab-6=-2b2-5ab+2(2)原式=3x2y-2x2y+6xy-3x2y+xy=-2x2y+7xy,当x=-1,y=-2时,原式=-2×(-1)2(-2)+7×(-1

(x3-2y3-3x2y)-(3x3-3y3-7x2y)

原式=x3-2y3-3x2y-3x3+3y3+7x2y=-2x3+y3+4x2y

极限.

lim(x->0)[f(3x)+f(-2x)]/tanx(0/0)=lim(x->0)[3f'(3x)-2f'(-2x)]/(secx)^2=3f'(0)-2f'(0)=f'(0)=1Ans:B

单项式:5x2y,-6x2y,34x

5x2y+(-6x2y)+34x2y=14x2y答:和是-14x2y.

已知A=x3-2y3+3x2y+xy2-3xy+4,B=y3-x3-4x2y-3xy-3xy2+3,C=y3+x2y+2

因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.

2(x2y+xy)-3(x2y+xy)-4x2y其中x=-2,y=12

原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.

已知xy<0,则x2y化简后为 ___ .

∵xy<0,由二次根式的有意义,得y>0,∴x<0,∴原式=x2y=-xy.

当x=2011,y=2012时,求代数式3x3-4x3y2+3x2y+2x2+4x3y2+2x2y-5x2-5x2y+x

化简得:9-12Y^2+6Y+4+12Y^2+4Y-10-10Y+X-Y+1=X-Y+4带入X、Y值得:=3

(X2 -y+1)(X2+1)+X2y+y -X2因式分解

(X^2-y+1)(X^2+1)+X^2y+y-X^2=(X^2-y+1)(X^2+1)+(X^2+1)y-X^2=(X^2-y+y+1)(X^2+1)-X^2=(X^2+1)^2-x^2=(x^2+

把多项式2xy2-x2y-x3y3-7按x作升幂排列是______.

根据题意得:-7+2xy2-x2y-x3y3.故答案为:-7+2xy2-x2y-x3y3

一个多项式加上3x2y-3xy2得x3-3x2y,则这个多项式是(  )

根据题意得:(x3-3x2y)-(3x2y-3xy2)=x3-3x2y-3x2y+3xy2=x3-6x2y+3xy2,故选C.

一个多项式加上-2x3-3x2y+5y2,得x3-2x2y+3y2.

(1)(x3-2x2y+3y2)-(-2x3-3x2y+5y2)=x3-2x2y+3y2+2x3+3x2y-5y2=3x3+x2y-2y2,答:这个多项式为3x3+x2y-2y2.(2)当x=-12,

4x2y-{x2y-「3xy2 – 1/ 2(4x2y-8xy2)+x2y」}-5xy2

答案:2x^2y+2xy^2原式=4x2y-{x2y-[3xy2-2x2y+4xy2+x2y]}-5xy2=4x2y-{x2y-[7xy2-x2y]}-5xy2=4x2y-{x2y-7xy+x2y]}

极限

e^(pi/n*∏ln(2+cosipi/n))指数是个积分公式=e^∫[0pi]ln(2+cosx)dx可以用参变积分求积分

一个多项式加上x2y-3xy2得2x2y-xy2,则这个多项式是(  )

(2x2y-xy2)-(x2y-3xy2)=2x2y-xy2-x2y+3xy2=x2y+2xy2.故选C.

x+xy+y=0 x3+x3y3+y3=12 三是次数、求解方程组

x^3+y^3+x^3y^3=12,x^3+y^3+x^3y^3+1=13,(x^3+1)(y^3+1)=13(x+1)(x^2-x+1)(y+1)(y^2-y+1)=13;x+y+xy=0,x+y+

单项式5x2y、3x2y、-4x2y的和为______.

5x2y+3x2y+(-4x2y)=(5+3-4)x2y=4x2y,故答案为:4x2y.

化简求值:2(x2y+xy)-3(x2y-xy)-4x2y,其中x=-1,y=1.

原式=2x2y+2xy-3x2y+3xy-4x2y=-5x2y+5xy,当x=-1,y=1时,原式=-5×(-1)2×1+5×(-1)×1=-5-5=-10.

因式分解:x3-y3-x2y+xy2

x3-y3-x2y+xy2=(x-y)(x2+xy+y2)-xy(x-y)=(x-y)(x2+xy+y2-xy)=(x-y)(x2+y2)