X (Y-X) 1, 2x-2y 1 If 1 of X + x = 5, then x & # 178; + X & # 178; 1 of x = (), x-x 1 of x = (), X & # 178; - 5x + 3 = ()

X (Y-X) 1, 2x-2y 1 If 1 of X + x = 5, then x & # 178; + X & # 178; 1 of x = (), x-x 1 of x = (), X & # 178; - 5x + 3 = ()

（1）
x+1/x=5
(x+1/x)²=5²
x²+1/x²+2=25
x²+1/x²=23
（2）
(x-1/x)²=x²+1/x²-2=23-2=21
x-1/x=±√21
（3）
x+1/x=5
x²+1=5x
x²-5x=-1
therefore
x²-5x+3=-1+3=2

To solve the equations: x + 2Y + 3Z = 80, X / 2 + Y / 3 = Z / 4,

x+2y+3z=80.1
,x/2+y/3=z/4.2
From formula 2, 6x + 4Y = 3Z is obtained
7x + 6y = 80 solve whether there is an array, such as:
x=2,y=11,z=56/3
x=4,y=26/3,z=176/9
x=8,y=4,z=64/3

Solving the system of linear equations with three variables {X / 3 = x / 4 = x / 5 x-2y + 3Z = 30

x/3=y/4=z/5
Let X / 3 = Y / 4 = Z / 5 = t
∴x=3t,y=4t,z=5t
Substituting x = 3T, y = 4T, z = 5T into (2), we get
3t-8t+15t=30
t=3
∴x=9
y=12
z=15

5x+2y-3z 3x-2y+2z=14 4x-2x-5z=9

The first is incomplete
Questioning

2X+Y+Z=7 X+2Y+3Z=14 2X+2Y+5Z=21

2X+Y+Z=7①
X+2Y+3Z=14②
2X+2Y+5Z=21③
③-①,y+4z=14
② * 2 - ①, 3Y + 5Z = 21, the solution is y = 2, z = 3
The solution is x = 1

x-2y+3z=0,3x+2y+5z=12,2x-4y-z=(-9)

x-2y+3z=0 （1）
3x+2y+5z=12 （2）
2x-4y-z=(-9) （3）
(1) Formula + (3) is as follows:
3x-6y+2z=-9 (4)
(2) Formula - (4) is as follows:
8y+3z=21 (5)
(1) The formula is multiplied by 3
3x-6y+9z=0 （6）
(2) Formula - (6) is as follows:
8y-4z=12 (7)
(5) Formula - (7) is as follows:
7z=9,z=9/7；
Substituting (7): 8y-36 / 7 = 12;
y=15/7；
Substituting z = 9 / 7; y = 15 / 7; into (1)
x-30/7+27/7=0
x=3/7；
So the solution is:
x=3/7；
y=15/7；
z=9/7

X: y = 4: 7, X: z = 3: 5, x minus 2Y plus 3Z = 30, how many are x, y and Z

x=12,y=21,z=20
x:y=4:7+12:21,X:z=3:5=12：20

If x is greater than or equal to 0, y is greater than or equal to 0, Z is greater than or equal to 0.3x + 2Y + Z = 5, 2x + y-3z = 1, u = 3x + y-7z, find the maximum value of U,

y=7-11z
x=7z-3
And X, y, Z ≥ 0
7 / 11 ≥ Z ≥ 3 / 7
And u = 3z-2
The maximum value of u is - 1 / 11 and the minimum value is - 5 / 7

X minus y minus Z equals 1 2x plus y minus 3Z equals 4.3x minus 2Y minus Z equals minus 1

X -- y -- z = 1 (1) 2x + Y -- 3Z = 4 (2) 3x -- 2Y -- z = -- 1 (3) (1) + (2): 3x -- 4Z = 5 (4) (3) - (1) x2: x + Z = -- 3 (5) (4) + (5) X4: 7x = -- 7

Given that X: Y: z = 4:5:6 and X + 2Y + 3Z = 64, we can find the value of XYZ

x:y：z=4:5:6
Let x = 4K, y = 5K, z = 6K
∵x+2y+3z=64
∴4k+10k+18k=64
32k=64
k=2
∴x=8
y=10,z=12