假设x=5,y=8 则z=x ( y)

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(x+y-z)(x-y+z)=

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai

(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=4x-2y-z-5x[6y-(x+y)]+x-(3y-10z)=4x-2y-z-30xy+5x²+5xy+x-3

2x+5y+4z=0,3x+y-7z=0,则x+y-z=?

解法2:2x+5y+4z=0式①3x+y-7z=0式②x+y-z=?式①×3-式②×23(2x+5y+4z)-2(3x+y-7z)=015y+12z-2y+14z=013y+26z=0式③式①-式②×

已知x+y=3,y+z=4,x+z=5,则x+y+z等于(  )

由题意得:x+y=3①y+z=4②x+z=5③①+②+③得:2x+2y+2z=12,即x+y+z=5.故选A.

已知方程组4x-5y+2z=0 x+2y=3z 则x:y:z

4x-5y+2z=0(1)x+2y=3z(2)(2)×4-(1)得:13y=14zy=14/13z(1)×2+(2)×5得:13x=11zx=11/13z所以:x:y:z=11/13:14/13:1=

已知2x+5y+4z=15.7x+y+3z=14 则x+y+z=?

设a(2x+5y+4z)+b(7x+y+3z)=x+y+z比较系数得2a+7b=5a+b=4a+3b=1a=1/11,b=2/11因此x+y+z=a(2x+5y+4z)+b(7x+y+3z)=1/11

2X+Y+Z=10,X+2Y+Z=-6,X+Y+2Z=8 5X+4Y-3Y=13,2X-3Y+7Z=19,3X-Y=2X

2x+y+z=10(1)x+2y+z=-6(2)x+y+z=8(3)(2)-(3):y=-14(4)(1)-(2):x-y=16(5)把(4)代入(5):x+14=16x=2(6)把(4)和(6)代入

若4x=7y+5z,2x+y=z,则x:y:z的值为

把z=2x+y代入4x=7y+5Z得4x=7y+10x+5yx=-2y,z=-3yx:y:z=-2:1:-3

x+y+z=51 4x+8y+5z=300 x+y+2z=67

(3)-(1)得,z=16代入(1)得,x+y=35(4)代入(2)得,4x+8y=220(5)(5)-4×(4)得,4y=80解得,y=20代入(4)得,x+20=35解得,x=15所以,方程组的解

假设有条件 int x=1,y=2,z=3;则表达式z+=x>y?++x:++y的值是6,为什么是6呢?

,比目运算符(?:)和>的优先级都比+=高.所以上式可以这样看z=z+(x>y?++x:++y)z=z+(++y)=3+(2+1)=6

假设有条件 int x=1,y=2,z=3;则表达式z+=x>y?++x:++y的值是【1】.6

x>y?++x:++y执行后的值为++y,而后y=3;z+=y;所以z=3+3=6再问:为什么不是先计算z+=x在判断z>y呢??

若3x+7y+z=5,4x+10y+z=3,则x+y+z=?

3x+7y+z=5.(1)4x+10y+z=3.(2)(1)*3-(2)*2得x+y+z=15-6=9所以x+y+z=9

{x+y+z=1;x+3y+7z=-1;z+5y+8z=-2

这个题目没有问题么,我是说最后一个式子确定是z+5y+8z=-2?如果没有问题的话:x+y+z=1;①x+3y+7z=-1;②z+5y+8z=-2③①-②2Y+6Z=-2Y=(-2-6Z)/2=-1-

X+Y+Z=?

X+Y+Z

f(x,y,z,w)=x*(x+y)*(x+y+z)*(x+y+z+w)

f=x+1f+u=2x+3f+u+c=3x+8f+u+c+k=4x+15f(f,u,c,k)=(x+1)(2x+3)(3x+8)(4x+15)

x/2=y/3=z/5 x+3y-z/x-3y+z

设x/2=y/3=z/5=ax=2ay=3az=5a是不是求的是:(x+3y-z)/(x-3y+z)?若是,如下:(x+3y-z)/(x-3y+z)=(2a+9a-5a)/(2a-9a+5a)=-3

若x+3y+5z=10,5x+z+3y=8,x+y+z=?

x+3y+5z=10,5x+z+3y=8两式相加6x+6y+6z=18x+y+z=3

若{x+3y+10z=0 则 (x+y-z)/(x-y+z)

x+3y+10z=0就是x+3y=-10z即2x+6y=-20zA式2x-y-2z=0就是2x-y=2zB式A式-B式得到:(2x+6y)-(2x-y)=-20z-2z即7y=-22z解出y=-22z