x的方-6x(y-z) 9(y-z)的方-搜答案-智慧作业帮

x+2y-9z=0x-2y-5z=0相加2x-14z=0x=7z相减4y-4z=0y=z所以原式=(98z²+3z²+7z²)/(49z²-4z²+9

x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1

(x+y+z)^3-x^3-y^3-z^3=(x+y+z)^3-x^3-(y^3+z^3)=(x+y+z-x)[(x+y+z)^2+x(x+y+z)+x^2]-(y+z)(y^2-yz+z^2)=(y

(x+y)^3+(z-x)^3-(y+z)^3=[(x+y)^3-(y+z)^3]+(z-x)^3=[(x+y)-(y+z)][(x+y)^2+(x+y)(y+z)+(y+z)^2]+(z-x)^3=

y=6-x所以z²=6x-x²+9(x-3)²+z²=0所以x-3=0,且z=0所以z=0

先转化到2个未知数用另外一个未知数表示,然后代入求值4x-3y-6z=0x+2y-7z=0解得x=3zy=2z（2x²+3y²+6z²)/(x²+5y²

(x+y+z)^2-(x-y-z)^2=(x+y+z+x-y-z)(x+y+z-x+y+z)=2x(2y+2z)=4x(y+z)=4xy+4xz

（X+Y+Z)²=X²+Y²+Z²+2(XY+YZ+XZ)X²+Y²+Z²=10²-2×8=84

(x²-6x+9)+(y²+2y+1)+|z+3|=0(x-3)²+(y+1)²+|z+3|=0所以x-3=y+1=z+3=0x=3,y=-1,z=-3所以x-

x^2-4x(y+z)^2+4(y+z)^2=[x-2(y+z)]^2=(x-2y-2z)^2

题目有问题吧,我感觉应该是(x²+y²-z²)²-4x²y²(x²+y²-z²)²-4x²

∵z为有理数∴z^2=xy-9>=0∴y(6-y)-9>=0y^2-6y+9

原式=24a²b³(x+y-z)+18a³bc²(x+y-z)-36abc(x+y-z)=6ab(x+y-z)(4ab²+3a²c²

2x+y-3z=1,①x-2y+z=6,②3x-y+2z=9③①+③得：5x-z=10④①×2+②得：5x-5z=8⑤④-⑤得：4z=2∴z=1/2x=21/10=2.1y=-1.7

x^2+y^2+z^2-2x+4y+6z+14=0x^2-2x+1+y^2+4y+4+z^2+6z+9=0(x-1)^2+(y+2)^2+(z+3)^2=0x=1y=-2z=-3

因为3x^2-y^2+2z^2,将x=1-3y-2z代入,那么26y^2-2(18z-9)y+(14z^2-12z-3x^2-y^2+2z^2-3)=04(18z-9)^2-4*26(14z^2-12

按照分式等号左边分子分母相消后只剩分母16的思路去解.在（16/9）的y次方里拿出个16来,剩下的东东让他们处理完得1.(注意15=5*38=2*2*29=3*316=2*2*2*227=3*3*31

-3X的5次方Z再问：确定吗再答：这可以确定的，只要你没给错题

有10件产品,其中8件正品,2件次品,甲乙先后各取一件,求甲先抽到正品条件下,已抽到正品的概率?这个有答案吗?

x+2y-9z=0,x-2y-5z=0两式相加,2x-14z=0x=7z两式相减,4y-4z=0y=z2x方+3y方+7z方∕x方-4y方+9z方=(98z²+3z²+7z