x=10,y=5,z=20,表达式 x
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[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)]=4x-2y-z-5x[6y-(x+y)]+x-(3y-10z)=4x-2y-z-30xy+5x²+5xy+x-3
(1)2x+3y-4z=-5(2)x+y+z=6(两边同时×33x+3y+3z=18(与(1)相减得(5)(3)x-y+3z=10(与(2)相加得(4))(4)2x+4z=16(5)x+7z=23(两
x+y+z=A(3x+7y+z)+B(4x+10y+z)易知3A+4B=1,7A+10B=1,A+B=1,解得A=3,B=-2,所以原式=3*5-2*6=3,以后也是这样做就好了!给分吧!
x+y+z=70①x/30+y/20+z/40=2/5②x/40+y/20+z/30=10/23③①-30*②得-1/2y+1/4z=58④①-40*③得-y-1/3z=1210/23⑤2*④-⑤得5
根据给的条件可以得出:X/16+Y/10+Z/20=2.5可化为5X+8Y+4Z=2.5*80(1式)Z/16+Y/10+X/20=2.65可化为5Z+8Y+4X=2.65*80(2式)(1式)-(2
93x+7y+z=5所以6x+14y+2z=10又因为4x+10y+z=3所以2x+4y+z=7原题中两式相减得x+3y=-2所以x+y+z=9
3x+7y+z=5.(1)4x+10y+z=3.(2)(1)*3-(2)*2有9x+21y+3z-(8x+20y+2z)=5*3-3*2x+y+z=15-6x+y+z=9
x+2y+3z=20.(1)x+3y+5z=31.(2)(1)*2-(2)得x+y+z=9
x+y=10y+z=15z+x=203式相加2x+2y+2z=10+15+20x+y+z=22.5x=22.5-15=7.5y=22.5-20=2.5z=22.5-10=12.5
x+y-z=51式2x+3y+z=102式x-2y-z=203式1式+2式得3x+4y=154式2式+3式得3x+y=305式4式-5式得y=-5将y=5代入5式得x=35/3则Z=-5/3
①+②+③6X+6Y+6Z=14X+Y+Z=7/3④①-④4X=11/3X=11/12③-④4Y=-13/3Y=-13/12③-④4Z=23/3Z=23/12
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
3X+4Y=10①4Y-2Z=20②5X+2Z=30③由于注意到②有-2z,③有+2z②③两式相加4y+5x=50④4y+3x=10①④-①:2x=40x=20代入①4y=-50y=-25/2将x=2
x=1y=2z=7
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
x+3y+5z=10,5x+z+3y=8两式相加6x+6y+6z=18x+y+z=3
解:不防设x=2A,则y=3A,z=5A.由x+y+z=20,可知2A+3A+5A=20,10A=20,A=2.则x=4,y=6,z=10.
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