1，加7精魄要多少幸运值
这样和你说吧，只有要加7要幸运值才不亏，幸运值只需要55点，加6用幸运值那就是亏了 53天魔为你解答
+5，+6，+7都是65点，谢谢
65吧。
2，解一下 yzu1 zux2 uxy3 xyz47
y+z+u=1 ①z+u+x=2 ②u+x+y=3 ③x+y+z=47 ④①+②+③+④得 3(x+y+z+u)=53∴x+y+z+u=53/3 ⑤⑤-①得 x=50/3⑤-②得 y=47/3⑤-③得 z=44/3⑤-④得 u=-88/3
因为x:y:z:u=1:2:3:4 所以y=2x,z=3x,u=4x 9x+7y+3z+2u=200 9x+14x+9x+8x=200 40x=200 x=5 所以y=10,z=15,u=20
3，已知xyzuyzuxzuxyuxyz求xyzuyzxuzux
由合分比定理得x/(y+z+u)=y/(z+u+x)=z/(u+x+y)=u/(x+y+z)=(x+y+z+u)/(3x+3y+3z+3u)=1/3∴3x=y+z+u....(1)3y=z+u+x....(2)3z=u+x+y....(3)3u=x+y+z....(4)(1)-(2),得3(x-y)=y-x∴x=y同理y=z, z=u∴x=y=z=u代入(x+y)/(z+u)+(y+z)/(x+u)+(z+u)/(x+y)+(u+x)/(y+z)=1+1+1+1=4
由合分比定理知x/(y+z+u)=y/(z+u+x)=z/(u+x+y)=u/(x+y+z)=(x+y+z+u)/(3x+3y+3z+3u)=1/3化简得四元一次方程组，仅三个独立，得x=y=z=u代入问题可得结果。(x+y)/(z+u)+(y+z)/(x+u)+(z+u)/(x+y)+(u+x)/(y+z)=2x/2x+2x/2x+2x/2x+2x/2x=4
y+z+u=1 ①z+u+x=2 ②u+x+y=3 ③x+y+z=47 ④①+②+③+④得 3(x+y+z+u)=53∴x+y+z+u=53/3 ⑤⑤-①得 x=50/3⑤-②得 y=47/3⑤-③得 z=44/3⑤-④得 u=-88/3
