4x² 18y² z²-12xy-6yz=0-搜答案-大师作文网

（一）x(x+y+z)=4-yz.===>x²+(y+z)x+yz=4.===>(x+y)(x+z)=4①.同理,将后面两个方程变形可得(x+y)(y+z)=9,②(x+z)(y+z)=25

x^4-46x²+25=x^4-10x²+25-36x²=(x²-5)²-36x²=(x²-5+6x)(x²-5-6x)

把x=6-y带入z^2-4z+4=xy-9中,得(y-3)^2+(z-2)^2=0,故y-3=0,z-2=0,所以y=3,z=2,x=3.

处理这类比例问题,有一个通用方法如果：x:y:z=a:b:c可以设x=aky=bkz=ck带入计算,就行了自己来试试吧~

题目应为:xy/(x+y)=6/5yz/(y+z)=12/7xz/(x+z)=4/3求x和y和z运用倒数变形可解因为1/y+1/x=5/6,1/z+1/y=7/12,1/z+1/x=3/4三式相加得1

z²-4z+4=xy-9又x=6-y,代入得z²-4z+4=（6-y）y-9(z-2)²=-(y-3)²(z-2)²+(y-3)²=0所以(

已知x^3+y^3-z^3=96,xyz=4,x^2+y^2+z^2-xy+xz+yz=12,则x+y-z等于[x+y-z]^2=x^2+y^2+z^2-2xy-2xz-2yzx^3+y^3=（x+y

X=1,Y=2,Z=3其实很简单!

xy\X+Y=12\71/y+1/x=7/12(1)YZ\Y+Z=6\51/z+1/y=5/6(2)XZ\X+Z=4\31/z+1/x=3/4(3)由(1)-(2)得1/x-1/z=-1/4(4)由(

2x+3y-5z=0,3x-2y+12z=0(z≠0),解得x=-2zy=3z所以2x*2-3xy/4x*2-12xy+9y*2=（8z²+18z²）/(16z²+72z

原式=(3/4x^3y^3z^3)×12/(xy)-(1/3x^2y^2z)×12/(xy)+(1/2xy)×12/(xy)=9x²y²z³-4xyz+6

只能提取公因式-16x²y²+12xy³z-4xy=-4xy(4xy+3y²z+1)再问：因式分解3a²-6a+3谢谢再答：=3(a²-2a

xy+xz+yz=76,x/3=y=z/4所以,19x^2/9=76,x^2=362x*x+12y*y+9z*z=58x^2/3=58*12=696

第二个分母写错了?(y-x)(z-x)/(x-2y+z)/(x+y-2z)+(z-y)(x-y)/(x+y-2z)/(y+z-2x)+(x-z)(y-z)/(y+z-2x)/(x-2y+z)=1

答案是：(2*X)/((X-Z)*(X+Z))再问：解题过程给我写下1再答：=（2X+Z-Y)/[(x-y)(x+z)]-(y-z)/[(x-z)(x-y)]=[(2x+z-y)(x-z)-(y-z)

3y=x+2y错了吧是不是3y=x+2z?x-3y+2z=0x²+9y²+4z²-6xy-12yz+4xy=(x²-6xy+9y²)-12yz+4xy

原式=(x-2y)^-z^=(x-2y-z)(x-2y+z)

x²-4xy+4y²-9z²

因式分解吧x²-4xy+4y²-9z²=（x-2y）^2-(3z)^2=(x-2y+3z)(x-2y-3z)

解题思路：本题的关键是将三个方程两边取倒数，化简后分别将方程等号左边和右边相加，得到1/x+1/y+1/z的值，最后将要求的分式化简，把1/x+1/y+1/z的值带入即可。解题过程：

3x-y=-2zx+2y=-3z那么：x=-z,y=-z（3x^－xy+2y^)/(2x^+4xy+y^)=(3z^2-z^2+2z^2)/(2z^2+4z^2+z^2)=4z^2/7z^2=4/7