x² 2y²-2xy-4y 4=0求x的y次方

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/18 21:06:19
分解因式1.ax-2ay+2bx-4by2.x^3+x^2-4x-43.x^2-x-4y^2+2y4.4xy+1-4x^

1.ax-2ay+2bx-4by=ax+2bx-(2ay+4by)=(x-2y)(a+2b)2.x^3+x^2-4x-4=x(x^2-4)+(x^2-4)=(x-1)(x-2)(x+2)3.x^2-x

已知2x-y=1/3 xy=2 求2x4次方y3次方-x3次方y4次方的值

原式分解因式得x^3y^3(2x-y)=(xy)^3(2x-y)=8/3.(x^3表示x的3次方)

设函数Y=f(x)由x2+3y4+x+2y=1所确定,求dy/dx

把原式两边对x求导得:x^2+12y^3*dy/dx+1+2dy/dx=0合并同类项移项得:dy/dx=-(1+2x)/(12y^3+2)

x,y are positive integers.2x+y4,greatest possible x-y?

题目翻译:x,y是正实数.2x+y < 29,且y > 4,求问x-y的结果最大的结果?解法:要求x-y最大的结果,在正实数范围内,那x应该尽量大,

(-3x^y+2xy)-( )=4x^+xy

(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^

解方程组:x3-y4=13x-4y=2

原方程组可化为4x-3y=12  ①3x-4y=2  ②,①×4-②×3,得7x=42,解得x=6.把x=6代入①,得y=4.所以方程组的解为x=6y=4.

已知:x+y=6,xy=4.(1)求x2+y2的值;(2)求(x-y)2的值;(3)求x4+y4的值.

∵x+y=6,xy=4,∴(1)x2+y2=(x+y)2-2xy,=62-2×4,=28;(2)(x-y)2=x2+y2-2xy,=28-2×4,=20;(3)x4+y4=(x2+y2)2-2x2y2

设函数y=f(x)由方程xy+2lnx=y4所确定,则曲线y=f(x)在点(1,1)处的切线方程是______.

等式xy+2lnx=y4两边直接对x求导,得y+xy′+2x=4y3y′将x=1,y=1代入上式,有 y'(1)=1 故过点(1,1)处的切线方程为y-1=1•(x-1),即x-y

已知:x2+y2=4xy,求(x4+y4)÷(xy)2

(x⁴+y⁴)÷(xy)²=[(x²+y²)²-2x²y²]/(x²y²)=[(4xy)

(15x^4y4-9x^5y^3-3x^6y^2)/(-3x^2y)^2

解(15x^4y^4-9x^5y³-3x^6y²)/(-3x²y)²=(15x^4y^4-9x^5y³-3x^6y²)/(3x²y

因式分解:x4次方-2x²y²+y4次方

原式=(x²-y²)²=(x+y)²(x-y)²

已知x+y+z=0,x2+y2+z2=1,求xy+yz+xz,x4+y4+z4的解

(x+y+z)^2=[(x+y)+z]^2=(x^2+2xy+y^2)+z^2+2zx+2zy=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2(xy+xz+yz)=0x+y

已知3x^2+xy-2y^2=0,求{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}

3x^2+xy-2y^2=0推出(3x-2y)(x+y)=0推出x=-y或x=(2/3)y{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}/x^2-9y^2推出:{

已知x*x-4xy+4y*y=0 求[2x(x+y)-y(x+y)]/(4x*x-4xy+y*y)的值?

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

正实数x,y满足xy=1,那么1x4+14y4的最小值为(  )

由已知,得x=1y,∴1x4+14y4=1x4+x44=(1x2-x22)2+1,当1x2=x22,即x=42时,1x4+14y4的值最小,最小值为1.故选C.

已知x+y=4 xy=1,求x2+y2,x3+y3,x4+y4,x5+y5 x6+y6 x7+y7

x2+y2=(x+y)2-2xy=14x3+y3=(x2+y2)×(x+y)-xy2-yx2=14×4-xy(x+y)=52……剩下的就是这么个算法,手机党,求个最佳哈

yy''-(y')2=y4满足当x=0时,y=1;当x=0时,y'=o

很明显的,不含x型,二阶微分方程,令y'=t,y''=t*t'y*t*t'-t^2=y^4然后t*t'=(t^2)'/2令t^2=s(y/2)*s'-s=y^4然后用公式解即可,解的过程就不用说了吧

已知x+y+z=0,求x4+y4+z4-2x2y-2y2z2-2z2x2的值

(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了

因式分解1)2x³y—4x平方y平方+2xy³ 2)x4次方—2x平方y平方+y4次方 3)3m(x

1)2x³y—4x平方y平方+2xy³=2xy(x²-2xy+y²)=2xy(x-y)²2)x4次方—2x平方y平方+y4次方=(x²-y&

已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x

因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²