(12x²y²z²-4xy²) 2xy
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(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
解题思路:本题主要利用完全平方公式进行因式分解即可求出结果解题过程:解:x²-6xy+9y²=(x-3y)2
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
解题思路:将原方程右边分解因式为(x-4y)(x+y)=0,可得x=4y或x=-y,进而可求出x与y的比值。解题过程:解:∵x2-3xy-4y2=0∴(x-4y)(x+y)=0∴x=4y或x=-y∵x
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
解题思路::∵x+y=0,x+13y=1,解得x=1/12,y=-1/12∴x²+12xy+13y²=1/144-1/12+13/144=14/144-1/12=2/144=1/72解题过程:已知x+
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
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)
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