{5x-2y + Z = 12 {4x + 3Y + Z = 1 {7x + 4Y + Z = 2 (solution, process) Detailed, I will increase the reward, thank you!

{5x-2y + Z = 12 {4x + 3Y + Z = 1 {7x + 4Y + Z = 2 (solution, process) Detailed, I will increase the reward, thank you!

5x-2y+z=12 ………… (1)
4x+3y+z=1………… II.
7x+4y+z=2………… 3.
①-②：x-5y=11………… 4.
①-③：-2x-6y=10………… 5.
④*2+⑤：-16y=32
So: y = - 2
Substituting (4): x = 1
So: z = 3
In order to factorize the quadratic trinomial x2-5x + P in the range of integers, the value of integer P can be () A. 2 b. 4 C. 6 D. innumerable
If the quadratic trinomial x2-5x + P can be decomposed, there must be: 25-4p ≥ 0, that is, P ≤ 254, which can be factorized within the range of integers. Therefore, as long as P can be decomposed into two integers, and the sum is - 5, such numbers have arrays, so there can be innumerable values of integer P. therefore, choose D
Y = 2x & # 178; - 6x + 1 x ∈ [1,3] for range
Y = 2 (x-3 / 2) & # - 7 / 2; its axis of symmetry is x = 3 / 2,
When x = 3 / 2, y has a minimum value of y = - 7 / 2
Because x = 3 is farther from the axis of symmetry x = 3 / 2 than x = 1
So when x = 3, y has a maximum y = 1
So the range is [- 7 / 2,1]
I hope my answer is helpful to you. Another brother's answer is a little suspect
y=2(x-3/2)^2-7/2
The minimum value is - 7 / 2
Take generation 1 and get - 3
Three generations have to go in
So the range is [- 7 / 2,1]
y=2x²-6x+1 =2(x-3/2)^2-7/2
Axis X = 3 / 2
ymin=y(3/2)=-7/2,ymax=y(3)=1
Range [- 7 / 2,1]
2x-3y + 5 = 0 3x + 2Y = 12 y = 2x 7x-3y = 4 y + 1 = one third x + 1 2x-3y = 1 equations
The first two are together
First: x = 2, y = 3, second: x = 4, y = 8, third: x = - 1, y = - 1
If x + X + 5m, which is twice of the quadratic trinomial, can be factorized in the range of real number, then what is m equal to?
Δ=1-40M≥0
M≤1/40
Y = (10x & # 178; - 6x + 1) / X & # 178; range
X belongs to (1 / 5,1 / 2]
Supplementary answer:
y=(10x^2-6x+1)/x^2
=10+(1-6x)/x^2=10+(1/x)^2-(6/x)=1+(1/x-3)^2
1/5
1 / X belongs to [2,5]
y=10-6/x+1/x^2
Take 1 / X as an independent variable, and the axis of symmetry of this function is 1 / x = 3
In [2,3] simple decrease and [3,5] simple increase.
So the range is (1,5)
y=((1/x)-3)^2+1
So the range is [2,5]
Let me give you a hint. If we can find out the range of 1 / X and take 1 / X in the original function as a whole, then the original problem is to find the range of a quadratic equation of one variable
It is known that XY satisfies the equations {x + 2Y = 3, X-Y = - 2. Find the value of X / (X-Y) square times X-Y / x + y + (2x + 2 / X-Y) - 2
From X-Y = - 2, X is 2 larger than y. from x + 2Y = 3, x + 2x + 2 = 3, 3x = 3-2, 3x = 1, x = 1 / 3, y = 7 / 3
In order to factorize the quadratic trinomial x2-5x + P in the range of integers, the value of integer P can be () A. 2 b. 4 C. 6 D. innumerable
If the quadratic trinomial x2-5x + P can be decomposed, there must be: 25-4p ≥ 0, that is, P ≤ 254, which can be factorized within the range of integers. Therefore, as long as P can be decomposed into two integers, and the sum is - 5, such numbers have arrays, so there can be innumerable values of integer P. therefore, choose D
Y = x + 2 / X & # 178; - 6x + 9 use the discriminant method to find the range. When X & # 178; - 6x + 9 = 0, the value of y should be taken from the range, right?
When y = x + 2 / X & # 178; - 6x + 9yx ^ 2-6yx + 9y = x + 2yx ^ 2 - (6y-1) x + 9y-2 = 0y = 0, X-2 = 0x = 2. Therefore, y = 0 exists in the range. (1) when Y0, the quadratic equation with respect to X has real roots, so the discriminant is greater than or equal to 0, that is: (6y-1) ^ 2-4y (9y-2) > = 036y ^ 2-12y + 1-36y ^ 2 + 8y > = 0-4y + 1 > =
To solve the equations: 3x + 4Y = 19x − y = 4
3x + 4Y = 19, ① x − y = 4, ② is transformed into, x = y + 4, ③ is substituted into formula ①, 3 (y + 4) + 4Y = 19, the solution is, y = 1, substituted into ②, x = 5, so the solution of the equations is: x = 5Y = 1