8a^6=( )^3 -x^3y^6=( )^3 4^3×(1/4)^3=( )^3=_________ 2^8×0.5^7=2×( )^7=________

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8a^6=( )^3 -x^3y^6=( )^3 4^3×(1/4)^3=( )^3=_________ 2^8×0.5^7=2×( )^7=________
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8a^6=( )^3 -x^3y^6=( )^3 4^3×(1/4)^3=( )^3=_________ 2^8×0.5^7=2×( )^7=________
8a^6=( )^3 -x^3y^6=( )^3 4^3×(1/4)^3=( )^3=_________ 2^8×0.5^7=2×( )^7=________

8a^6=( )^3 -x^3y^6=( )^3 4^3×(1/4)^3=( )^3=_________ 2^8×0.5^7=2×( )^7=________
8a^6=( 2a²)^3
-x^3y^6=(-xy² )^3
4^3×(1/4)^3=( 4×1/4)^3=____1_____
2^8×0.5^7=2×(2×0.5 )^7=__2______

【2(x-y)^3-8(x-y)^2(x+y)+6y(x-y)^2】/2(x-y)^2= (1)(x+y)^2-(x+y)^3(2)(x-y)-(x-y)^3(3)(x-y)^2-(y-x)^3(4)2(x-y)-3(y-x)^2(5)4ab(a=b)^2-6a^2b(a+b)(6)(x+y)^2(x-y)+(x+y)(x-y)^2(7)2a(a-3)^2-6a^2(3-a)+8a(a-3)(8)24xy^2z^2(x+y-z)-32xyz(z-x-y)^2+8xy(z-x-y)(9)(a-b)(b-a)^n(b-a)^2n-1 [两解)(10)2^2=3,2^b 初一因式分解练习题,谁好心帮我解下表示平方1.2X~-3XY-X2.(X+Y)~(X+Y)(X-Y)(X+Y)(Y+Z)3.3X(a+2b)~-6XY(a+2b)4.a(x-y)-b(X-Y)-C(Y-X)5.(b-a)~-2a+2b6.(x-y)~-m(y-x)+y-x7.x(x-y)(a-b)-y(y-x)(b-a)8.已知a+b=13,ab=40,求 b+ab~的值 1、(-y^3)^5/y^3/ y^2+2(-y)^102、(x-y)^7/(y-x)^6+(x+y)^3/(x+y)^23、已知X^a=5,x^b=3,求X^(3a-2b)的值 如果A={(x,y)|x,y+2>0},B={(x,y)|2x+3y-6>0},C={(x,y)|x-4 已知集合A={y|y=x*x-6x+5},B={y|y=6x+3-9x*x}求A交B 几道 数学题急啊 对了再加20分(x+y)^2-(x+y)^3(x-y)^2-(x-y)^3 (x-y)^2-(y-x)^32(x-y)-3(y-x)^24ab(a+b)^2-6a^2b(a+b)(x+y)(x-y)-(x+y)^28a(x-y)^2-4b(y-x)a=-5 a+b+c=-5.2求代数式a^2(-b-c)-3.2a(c+b)的值 a) 5x-y=6 2x+y=1 b)3x-8y=2 2x+2y=-6 分式~会滴来哦1..-a^3/2c÷5a^2/-6cd2.y-3/4y-8÷(y+2-5/y-2)3.1+(a-1/1-a)÷(1+a/a^2-2a+1)4.(x/x-y-1)÷y^2/x+y×(x-x^2/x+y),其中,x=-2,y=-15.x+2/x-2=x/x+1 集合A={(x,y)|x-2y=1}B={(x,y)|x+3y=6},则A∩B= A∪B=? A={y|y=-4x+6},B={x|y=5x-3},求A∩B. 2x^2y*A=(-6x^3y^2+9X^2y^3 ,求A=? 如果(2x^2)y*A=(6x^2)(y^2)-(4x^3)(y^3),求多项式A 已知方程组x+2y=6,x-y=a-3a的解满足x>0,y 设A={(x,y)}|3x+2y|.设A={(x,y)}|3x+2y|.B={(x,y)|x-y=2|},C={(x,y)|2x-2y|=3},D={(x,y)|6x+4y=2|},求A∩B,B∩C,A∩D设A={(x,y)|3x+2y=1}.B={(x,y)|x-y=2},C={(x,y)|2x-2y=3},D={(x,y)|6x+4y=2},求A∩B,B∩C,A∩D a^6x=3,a^2y=2,求a^3x,a^2x+y的值 下列因式分解不正确的是?A.(x+3)²-1=(x+2)(x+4)B.-x^4+y²=(y+x²)(y-x²)C.2x²y-8y³=2y(x²-4y²)D.x^6-16x²=x²(x²+4)(x+2)(x-2) 设A={(x,y)|2x-y=1},B={(x,y)|5x+y=6},C={(x,y)|2x=y+1}(1)A∩B,(2)C∩D,(3)A∩(CUD)D={(x,y)|2x-y=8