-The square of x-11x-2 = 0

-The square of x-11x-2 = 0

-The square of X - 11x-2 = 0, the flat of X + 11x + 2 = 0, x = (- 11 ± root sign (11 ^ 2-4 × 2)) / 2, X1 = (- 11 + root sign 113) / 2, X2 = (- 11 - root sign 113) / 2
Cube of X - square of 6x + 11x-6
Cube of X - square of 6x + 11x-6
=x^3-6x^2+9x+2x-6
=(x^3-6x+9x)+(2x-6)
=x(x^2-6x+9)+2(x-3)
=x(x-3)^2+2(x-3)
=(x-3)(x^2-3x+2)
=(x-3)(x-1)(x-2)
11x-7 = 26 to solve the equation
11X-7=26
11X=7+26
11X=33
X=33/11
X=3
11x + 5 = 25
11x=25-5
11x=20
x=20/11
11x=25-5
11x=20
We get x = 20 / 11
twenty-elevenths
11X+5=25
11X=25-5
11X=20
X=20÷11
X=20/11
11x=20
x=20/11
Given that X and y satisfy the following conditions: (1) x ≥ 0; (2) x + 2Y ≥ 3; (3) 2x + y is less than or equal to 3, find the minimum value of Z = X-Y
It's a piecewise function. Please tell me if you don't understand
minZ=-3
②x+2y≥3
③2x+y≤3
It can be deduced from the collation of (2) and (3)
(3-x)/2≤y≤3-2x
It can be seen from (1) that - x ≤ 0, so there is
y≤3+2*0=3
y≤3
-y≥-3
So, Minz ≥ 0 + (- 3) = - 3
When y = 3, Z is the minimum
When x is greater than or equal to 2, the minimum value of quadratic function y = 2x ^ 2-4x + 1
y=2(x-1)^2-1
∵x≥2
The minimum value of Y is 2 (2-1) ^ 2-1 = 1
The symmetry axis is x = 1, and the minimum value of x = 2 is f (2) = 1
If the length to width ratio of a rectangle is 4:3 and its diagonal length is 5 times the root sign 3, then the area of the rectangle is?
The area of this rectangle is 36
Let the variables X and y satisfy | x | + | y | ≤ 1, then the maximum and minimum values of X + 2Y are ()
A. 1，-1B. 2，-2C. 1，-2D. 2，-1
The constraint condition | x | + | y | ≤ 1 can be reduced to: x + y = 1, X ≥ 0, y ≥ 0x − y = 1, X ≥ 0, y < 0 − x + y = 1, x < 0, y ≥ 0 − x − y = 1, X < 0, y < 0. The plane area represented by it is shown in the figure below: from the figure, when x = 0, y = 1, x + 2Y takes the maximum value 2, when x = 0, y = - 1, x + 2Y takes the minimum value - 2, so select B
Write out the maximum and minimum values of quadratic function f (x) = (x-m) ^ 2 in the interval [- 1,1]
This paper discusses (1) if M ≤ - 1, then the axis of symmetry is on the left side of - 1, so f (x) increases on [- 1,1], then the maximum value is f (1) = (1-m) & sup2;, and the minimum value is f (- 1) = (1 + m) & sup2;; (2) if - 1 < m < 0, then the axis of symmetry is between [- 1,1], and f (1) is far away from the axis of symmetry, so the maximum value is f (1) = (1-m) & sup2
Axis of symmetry x = m, opening upward
When x < m, it decreases monotonically; when x > m, it increases monotonically
(1)
If M ＜ - 1, then the interval [- 1,1] is on the right side of the symmetry axis and increases monotonically
Minimum f (- 1) = (- 1-m) ^ 2 = (1 + m) ^ 2
Maximum f (1) = (1-m) ^ 2
（2） If M > 1, then the interval [- 1,1] is on the left side of the axis of symmetry and decreases monotonically
Maximum f (- 1) = (- 1-m) ^ 2 = (1 + m) ^ 2
Minimum f... expansion
Axis of symmetry x = m, opening upward
When x < m, it decreases monotonically; when x > m, it increases monotonically
(1)
If M ＜ - 1, then the interval [- 1,1] is on the right side of the symmetry axis and increases monotonically
Minimum f (- 1) = (- 1-m) ^ 2 = (1 + m) ^ 2
Maximum f (1) = (1-m) ^ 2
（2） If M > 1, then the interval [- 1,1] is on the left side of the axis of symmetry and decreases monotonically
Maximum f (- 1) = (- 1-m) ^ 2 = (1 + m) ^ 2
Minimum f (1) = (1-m) ^ 2
（3） If - 1 ＜ m ＜ 1, the axis of symmetry is in the interval [- 1,1], the minimum value is the minimum value
Minimum f (m) = (M-M) ^ 2 = 0
The larger of F (- 1) and f (1) is the maximum
f(-1)=(-1-m)^2=(1+m)^2
F (1) = (1-m) ^ 2
y=(x-m)²，x∈[-1，1]
① When m ≤ - 1, the function y is an increasing function on [- 1,1],
When x = - 1, y has the minimum value f (- 1) = (M + 1) & sup2;;
When x = 1, y has the maximum f (1) = (m-1) & sup2;;
② When - 1
Given that the two sides of a rectangle are a = 2 times the root sign 3 + 3, B = 2 times the root sign 3-3, find the diagonal of the rectangle
Diagonals = sum of squares of a + squares of B
Double root 6
(if the title you give is like this. Maybe the number you give B is wrong, and the root of 2 times 3-3 is equal to 0)
Zero