y=x^a a^x x^x

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y=x^a a^x x^x
(x-y)(x+y)(xx-yy)=?

(x-y)(x+y)(xx-yy)=(x^2-y^2)(x^2-y^2)=x^4-2x^2y^2+y^4

x+xx+xxx=xxxx

随便给个答案3+84+975=1062

已知14(XX+YY+ZZ)=(X+2Y+3Z)^2,求X:Y:Z

X:Y:Z=1:2:3因为:14(XX+YY+ZZ)=(X+2Y+3Z)^214(XX+YY+ZZ)-(X+2Y+3Z)^2=013X^2+10Y^2+5Z^2-4XY-6XZ-12YZ=0(4X^2

已知xx+y=5,xy=-3,则x^2y+xy^2=

x+y=5,xy=-3,则x^2y+xy^2=xy(x+y)=-3*5=-15

XX-YY/(X-Y)(X-Y)约分

XX-YY/(X-Y)(X-Y)=(x^2-y^2)/(x-y)(x-y)=(x+y)(x-y)/(x-y)(x-y)=(x+y)/(x-y)

已知2x=3y,求xy/xx+yy-yy/xx-yy的值

2/9再问:过程,谢谢再答:由题目得y/x=2/3xy/xx+yy-yy/xx-yy=y/x-(y/x)²=2/3-4/9=2/9

((xx+yy)/y-2y)/(1/y-1/x)^(1/(xx-yy))

原式=[(x²+y²-2y²)/y]/[(x-y)/xy]×1/(x²-y²)=[(x²-y²)/y]×[xy/(x-y)]×1/

已知函数f(x)=-4xx+4ax-4a-aa的定义域为A={x|xx≤x},f(x)的最大值为-5,求a的值

f(x)=-4x^2+4ax-4a-a^2A={x|x^2定义域为[0,1]f(x)=-4x^2+4ax-4a-a^2=-4(x^2-ax)-4a-a^2=-4(x-a/2)^2-4a->d=a/2(

已知xx+yy+4x-6y+13=0,求(xx-2x)/xx+3yy的值.

xx+yy+4x-6y+13=0整理得:(x+2)^2+(y-3)^2=0那么只有(x+2)=0(y-3)=0x=-2y=3(x^2-2x)/(x^2+3y^2)=(4+4)/(4+3*9)=8/31

已知x(x+1)-(xx+y)=3,求(xx+yy)/2-xy的值

x^2+x-x^2-y=3x-y=3(x-y)^2=9x^2+y^2-2xy=9(x^2+y^2)/2-xy=9/2

已知函数f(x)=xx-2x+3,g(x)=xx,则函数y=f[g(x)]的单调增区间是

f[g(x)]为复合函数,单调增区间,为f(x),g(x)单调性相同的区间;即同增,同减;f(x)=x^2-2x+3=(x-1)^2+2;x≥1;单调递增g(x)=x^2;x≥0;单调递增所以f[g(

已知关于x的方程xx-2ax+aa-2a+2=0的两个实数根x1 x2,满足x1x1+x2x2=

由韦达定理得:x1+x2=2ax1x2=a^2-2a+2因此有:x1^2+x2^2=(x1+x2)^2-2x1x2=4a^2-2a^2+4a-4=2a^2+4a-4=2即a^2+2a-3=0(a+3)

已知(x+y)(x+y)=25,(x-y)(x-y)=9求xy与xx+yy的值

(x+y)(x+y)=25x^2+2xy+y^2=25……(1)(x-y)(x-y)=9x^2-2xy+y^2=9……(2)(1)+(2)得:2x^2+2y^2=34x^2+y^2=17(1)-(2)

求使x的方程(a+1)xx-(aa+1)x+2aaa-6=0有整数根的所有整数a

是(a+1)x^2-(a^2+1)x+2a^3-6=0吗求△=(a^2+1)^2-4(a+1)*(2a^3-6)≥0-7a^4-8a^3+2a^2+24a+25≥07a^4+8a^3-2a^2-24a

已知 xx-2x+yy+6y+10=0 求x+y的值

x^2-2x+y^2+6y+10=0(x-1)^2+(y+3)^2=0所以x=1,y=-3x+y=-2x^2表示x的平方

已知XX+YY+4X-6Y+13=0,则X的Y次方是

XX+YY+4X-6Y+13=0(X+2)²+(Y-3)²=0X+2=0Y-3=0X=-2Y=3X的Y次方=-8

x(x+1)-(xx+y)=-3,求(xx+yy)/2-xy的值

x(x+1)-(xx+y)=-3x^2+x-x^2-y=-3x-y=-3(xx+yy)/2-xy=(x^2+y^2-2xy)/2=(x-y)^2/2=(-3)^2/2=9/2再问:是对的吧!再答:当然

xx-yy-x-y 因式分解

xx-yy-x-y=(x^2-y^2)-(x+y)=(x+y)(x-y)-(x+y)=(x+y)(x-y-1)

x y x+yy x+y xx+y x y

把所有列都加至第一列,第一列都是2x+2y将2x+2y提出,第一列剩下都是1,此时式外边有一因子(2x+2y)用2,3行加第1行负一倍得1yx+y0xy0x-y-x第一列展开得1*(-x^2-y(x-