[x+y/2=z+x/3=y+z/4,x+y+z=27]要有过程
- 教育综合
- 2024-11-30 12:59:59
[x+y/2=z+x/3=y+z/4,x+y+z=27]要有过程
12x+6y=12z+4x=12y+3z分成两个等式:
x+y/2 = y+z/4
=>4x+2y= 4y+z
=>4x=2y+z(一)
x+y/2=z+x/3
=>6x+3y = 6z +2x
=>4x = 6z-3y(二)
结合(一)(二)
6z-3y = 2y+z
5z = 5y
z = y
代入(二)
4x = y ×6 - 3y
4x = 3y
x = 3/4y
3/4 y + y + y = 27
3y + 4y + 4y = 108
11y = 108
y = 108/11
z = 108/11
x = 81/11
x+y/2=z+x/3=y+z/4 x+y+z=27
x+y/2=z+x/3=y+z/4 ①x+y+z=27 ②
①×12得:12x+6y=12z+4x=12y+3z
故:
12x+6y=12z+4x ③
12z+4x=12y+3z ④
由③得 x=(6z-3y)/4
把 x=(6z-3y)/4代入②和④中:
(6z-3y)/4+y+z=27
12z+4×(6z-3y)/4=12y+3z
解上述二元一次方程组得:y=z=108/11
将y=z=108/11代入②中得:x=81/11
∴x=81/11
y=108/11
z=108/11
x+y/2=z+x/3+y=z/4,...x+y+z=27.解方程组. X+Y/2=Z+X/3=Y+Z/4,X+Y+Z=27.解方程
设(x+y)/2=(x+z)/3=(y+z)/4=k则x+y=2k x+z=3k y+z=4k 代入x+y+z=272k+3k+4k=27 k=3所以 x+y= 6 (1)x+z=9 (2)y+z=12 (3)(1)+(2)+(3) 2(x+y+z)=27 x+y+z=27/2 (4)(4)-(1) z=15/2(4)-(2) y=9/2(4)-(3) x=3/2即为所求...xy/2=z+x/3=y+z/4 ,x+y+z=27 方程解 ,要过程
设x+y/2=z+x/3=y+z/4=k所以
x+y=2k
z+x=3k
y+z=4k
所以3个式子相加
2(x+y+z)=9k
因为,x+y+z=27
54=9k
k=6
所以
x+y=12
z+x=18
y+z=24
所以
z=27-12=15
y=27-18=9
x=27-24=3
祝学习进步
x+y/2=z+x/3+y=z/4,...x+y+z=27.解方程组。。。请写过程
设(x+y)/2=(x+z)/3=(y+z)/4=k则x+y=2k x+z=3k y+z=4k
代入x+y+z=27
2k+3k+4k=27 k=3
所以 x+y= 6 (1)
x+z=9 (2)
y+z=12 (3)
(1)+(2)+(3) 2(x+y+z)=27 x+y+z=27/2 (4)
(4)-(1) z=15/2
(4)-(2) y=9/2
(4)-(3) x=3/2
即为所求
展开全文阅读