若5x=125y,3y=9z,则x:y:z等于6:2:1类似的题?
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设x=3k,y=5k,z=7k.∵3x+2y-4z=9,∴9k+10k-28k=9,解得k=-1∴x=-3,y=-5,z=-7∴x+y+z=-3-5-7=-15.故选D.
(1)因为2x+3y-4z=7(1)3x-2y+z=3(2)x+5y+9z=20(3)(1)+(2)+(3)得(2x+3y-4z)+(3x-2y+z)+(x+5y+9z)=7+3+20整理得:6x+6
x+2y-9z=0,——(1)x-2y-5z=0——(2)(1)+(2)2x-14z=0x=7z将x=7z代入(1)7z+2y-9z=0y=z(2x^2+3y^2+7z^2)/(x^2-4y^2+4z
设x/3=y/4=z/5=m则x=3m,y=4m,z=5m则x+y+z/3x-2y+z=(3m+4m+5m)/(9m-8m+5m)=12/6=2
因为x:y:z=3:4:5所以设x=3k,y=4k,z=5k(k≠0)(1)z/(x+y)=5k/(3k+4k)=5k/7k=5/7(2)x+y+z=63k+4k+5k=612k=6k=1/2x=3k
设x/10=y/8=z/9=KX=10K,Y=8K,Z=9K(X+2Y+3Z)/(Y-5Z)=(10K+16K+27K)/(8K-45K)=-53/37
题目设置挺好的不会很变态,没什么难度由4x-3y-6z=0,x+2y-7z=0,可以解得x=3z,y=2z,将它代入代数式5x*5x+2y*2y-z*z/2x*2x-3y*3y-10z*10z=(25
再答:2016次方,和绝对值里面不可能为负,所以只能所有的都等于0
xy+yz+zx=93中的y,z全用x代替可以得到2x^2/3+10x^2/9+5x^2/3=93∴x^2=27同理y^2=12z^2=75∴9x*x+12y*y+2z*z=9*27+12*12+2*
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
2x+5y-4z=0,.(1)3x+9y-7z=0.(2)(2)-(1),得x+4y-3z=0.(3)(1)-(3),得x+y-z=0所以x+y-z的值为0
解前两个方程把X消掉求出Y=7-11Z因为Y≥0所以7-11ZY≥0所以Z≤7/11
x:y:z=3:4:5=6,求x-y+z分之x+y-z的值3y=4x,y=4x/3,3z=5x,z=5x/3x-y+z分之x+y-z=(x-4x/3+5x/3)/(x+4x/3-5x/3)=(4x/3
已知条件x:y:z=3:4:5且x+y+z=24设各自为x=3ny=4nz=5nx+y+z=24所以得出,3n+4n+5n=2412n=24n=2因此,x=3×2=6y=4×2=8z=5×2=10so
因为他们之间有比值关系,设x=3a,则:y=4a,z=5a三项相加等于24,3a+4a+5a=2412a=24a=2所以:x=3a=6,y=4a=8,z=5a=10
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
设2分之x+y=3分之y+z=5分之x+z=k则x+y=2k,y+z=3k,x+z=5k3式相加得2(x+y+z)=10k=18*2=36,k=3.6,x+y+z=5kz=5k-2k=3k=10.8x
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