2x²y²-x4一y4
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![2x²y²-x4一y4](/uploads/image/f/262632-48-2.jpg?t=2x%C2%B2y%C2%B2-x4%E4%B8%80y4)
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)
(x-y)(x+y)(x2+y2)(x4+y4)(x8+y8),=(x2-y2)(x2+y2)(x4+y4)(x8+y8),=(x4-y4)(x4+y4)(x8+y8),=(x8-y8)(x8+y8)
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
(1)原式=[(c-b-d)+a][(c-b-d)-a]=(c-b-d)2-a2=c2+b2+d2+2bd-2bc-2cd-a2,(2)∵x4-8x2y2+16y4=(x2-4y2)2∴原式=(x2-
原式=(x²-y²)²=(x+y)²(x-y)²
(x+y+z)²-(x²+y²+z²)=2(xy+yz+zx)=-1,xy+yz+zx=-1/2x3+y3+z3=3xyz+(x+y+z)(x²+y&
(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
∵x+y=a∴x2+y2+2xy=a2又∵x2+y2=b2∴2xy=a2-b2x4+y4=(x2+y2)2-2x2y2=(x2+y2)2-(2xy)22=b4−(a2−b2)22=-12a4+a2b2
解题思路:用平方差公式分解因式。解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/r
20x2+253x+800=(5X+32)(4X+25)(x+y)²=(x+y)²(x+y)²=(x²+y²+2xy)²=(x²+
原式=x4+x3y+4x3y+x2y+4x2y2+4x2y2+xy2+4xy3+xy3+y4,=x3(x+y)+4x2y(x+y)+xy(x+y)+4xy2(x+y)+y3(x+y),=-x3-4x2
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.
4分之x4次方y8次方-6x平方y平方+36y4次方=(1/2x²y的4次方-6y²)²如果本题有什么不明白可以追问,
原式=(x^4-2x²y²+y^4)+6xy(x²+2xy+y²)-2xy(x+y)=(x²-y²)²+6xy(x+y)²
(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了
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&
按x得降幂排列:x^4-4x^3y-x²y²+3xy^3-y^4按y得升幂排列:x^4-4x^3y-x²y²+3xy^3-y^4
x4-xy3-x3y-3x2y+3xy2+y4=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-
(x^4-y^4)÷(x^2+y^2)/(x+y)=(x^2-y^2)(x^2+y^2)÷(x^2+y^2)/(x+y)=(x^2-y^2)(x^2+y^2)*(x+y)/(x^2+y^2)=(x^2
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²