8(x y)的三次方 1
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2x^4*y^3-x^3y^4=x^3y^3(2x-y)=2^3*8=64
/>根据非负数之和为0得x+1=02(y-2)=0解得x=-1y=2于是x³-3xy+y³=-1³-3×(-1)×2+2³=-1+6+8=13
(-2/3x³y+4x²y²-1/2xy³)/(-xy/4)=-2/3x³y/(-xy/4)+4x²y²/(-xy/4)-1/2x
原式=(x²y²-1)(x²y²+1)-(x²y²-2xy+1)+4x²y²=x^4y^4-1-x²y&sup
(-8x²y³)÷(-2xy)²=(-8x²y³)÷(4x²y²)=-2y
o.5x的三次方y=|x|√2xy/2根号8xy的三次方=2|y|√2xy所以与√2xy都是同类二次根式
已知x+y=1,则x^3+y^3+3xy=(x+y)(x^2-xy+y^2)+3xy=1×(x^2-xy+y^2)+3xy=x^2-xy+y^2+3xy=x^2+2xy+y^2=(x+y)^2=1^2
(2x^4)y³-x³y^4=(x³y³)(2x-y)=(xy)³(2x-y)=4³×(1/16)=4
原式=xy(x²-2xy+y²)这一步就很清楚的明白是公式的运用了:a²-2ab+b²=(a-b)²所以=xy(x-y)²将x-y=-1,x
1.x³-2x²y+xy²=x(x²-2xy+y²)=x(x-y)²2.8a²b-2b=2b(4a²-1)=2b(2a-
1/2(a+2b-3c)+5(2/3a-1/2b+2/3c)-3(a-2b+c)=1/2a+b-3/2c+10/3a-5/2b+10/3c-3a+6b-3c=(1/2+10/3-3)a+(1-5/2+
首先x3-y3=(x-y)(x2+xy+y2)其次x3-y3-3xy=1所以x3-y3=1+3xy=(x-y)(x2+xy+y2)x-y=(1+3xy)/(x2+xy+y2)=(1+3xy)/[(x-
=xy(4x²+4xy+y²)=xy(2x+y)²
(-1/2xy)³=(-1/2)³x³y³=-1/8x³y³
x³y-xy³=xy(x²-y²)=xy(x+y)(x-y)
x³+y³-1+3xy=(x+y)(x²-xy+y²)+3xy-1=(x+y)[(x+y)²-3xy]+3xy-1=(x+y)³-3xy(x
(-xy)的五次方除以(-xy)的三次方=(-xy)的5-3次方=(-xy)²=x²y²再问:不好意思,是(-xy)的五次方除以(-xy)的二次方再答:=-x³
x+y=1,xy=3有问题,已不在实数范围内二分之一a²-0.5ab=a/2*[a-b]2009²-2008*2010=2009²-[2009-1]*[2009+1]=2
(3xy)³+(xy)³=27x³y³+x³y³=28x³y³很高兴为您解答,【the1990】团队为您答题.请点击下面