Let a, B and C be a three digit number of hundred, ten and one respectively, and a ≤ B ≤ C, then the maximum value that | A-B | + | B-C | + | C-A | can obtain is______ .

Let a, B and C be a three digit number of hundred, ten and one respectively, and a ≤ B ≤ C, then the maximum value that | A-B | + | B-C | + | C-A | can obtain is______ .

∵ a, B and C are three digit hundred, ten and one digit respectively, and a ≤ B ≤ C, ∵ a minimum is 1, C maximum is 9, ∵ A-B | + | B-C | + | C-A | = B-A + C-B + C-A = 2c-2a, ∵ A-B | + | B-C | + | C-A | the possible maximum value is 2 × 9-2 × 1 = 16
A three digit hundred digit number, ten digit number, and one digit number are a, B, and C (c > A) in turn. Exchange the position between the hundred digit number and one digit number, and the difference between the three digit number and the original three digit number is, it must be divisible
The original three digits are 100A + 10B + C
After the hundred digit and the individual digit exchange position, the three digit is 100C + 10B + a
The difference between the three digits and the original three digits is 99c-99a
It must be divisible by 99
Let a, B and C be a three digit number of hundred, ten and one respectively, and a ≤ B ≤ C, then the maximum value that | A-B | + | B-C | + | C-A | can obtain is______ .
∵ a, B and C are three digit hundred, ten and one digit respectively, and a ≤ B ≤ C, ∵ a minimum is 1, C maximum is 9, ∵ A-B | + | B-C | + | C-A | = B-A + C-B + C-A = 2c-2a, ∵ A-B | + | B-C | + | C-A | the possible maximum value is 2 × 9-2 × 1 = 16
If 3sina + cosa = 0, then one third of (COSA's square + sin2a) is much better
Because 3sina cosa = 0, so 3sina = cosa, so there is the square of sina plus the square of cosa equal to 1, and the square of sina equals 1 / 10
If the function f (x) = x2 + 2 (A-1) x + 2 is a decreasing function in the interval (- ∞, 4], then the value range of real number a is______ .
The axis of symmetry of the function f (x) = x2 + 2 (A-1) x + 2 is x = - 2 (a − 1) 2 × 1 = 1-A, and the function is a decreasing function in the interval (- ∞, 4). We can get 1-A ≥ 4 and a ≤ - 3
(1) Simplify y = cos ^ 4x-sin ^ 4x + 1 I know the last question is 2cos2x + 1. Can you simplify it? (2) y = 5 / 4 sin2x-5 / 2cos & # 178; X + 6
(1) Simplify y = cos ^ 4x-sin ^ 4x + 1 I know the last question is 2cos2x + 1. Can you simplify it?
(2)y=5/4 sin2x-5/2cos²x+6
1、y=(cos²x＋sin²x)(cos²x－sin²)＋1=cos²x－sin²x＋1=cos2x＋1===2cos²x－1＋1=2cos²x.2、y=(5/4)sin2x－(5/2)cos²x＋6=(5/4)sin2x－(5/4)[cos2x＋1]＋6=(5/4)[sin2...
The procedure specification specifies that the quadratic function f (x) = x & # 178; + 2 (A-1) x + 2 is a decreasing function in the interval (- ∞, 4], then the value range of real number a is_____
Tang Weigong's idea is correct, but it is short of the case of 1-A > = 4, so a
f(x)=x²+2(a-1)x+2 = f(x)=(x²+ a-1)² +2 - (a-1)²
The axis of symmetry is x = 1 - A
If f (x) is a decreasing function in the interval (- ∞, 4), the axis of symmetry can be on the right side of x = 4, that is, 1-A > 4, a < - 3
Finding the period of function sin ^ 4x + cos ^ 4x-2cos2x
What about the minimum and maximum?
sin^4 x+cos^4 x-2cos2x=(sin^2 x+cos^2 x)^2-2*sin^2x*cos^2x-2cos2x=1-(sin(2x)^2)/2-2*cos(2x)
=3/4+1/4*cos(4x)-2*cos(2x)
Because the periods of y = cos (4x) and y = cos (2x) are Pai / 2 and Pai respectively, and the ratio of the two periods is a rational number, the period of the original function is Pai / 2 and the least common multiple of Pai
When seeking the maximum value, y = (1 / 2) * cos ^ 2 (2x) - 2cos (2x) + 1 / 2 = (1 / 2) * (COS (2x) - 2) ^ 2-3 / 2, when cos2x = 1, y has the minimum value of - 1, when cos2x = - 1, y has the maximum value of 3
Solution: sin ^ 4x + cos ^ 4x-2cos2x = (sin ^ 2x + cos ^ 2x) ^ 2-2 * sin ^ 2x * cos ^ 2x-2cos2x = 1 - (sin (2x) ^ 2) / 2-2 * cos (2x)
=3/4+1/4*cos(4*x)-2*cos(2*x)
So the period is 2pi / 2 = Pi;
If the quadratic function f (x) = x & # 178; - (A-1) x + 5 is a monotone function in the interval {1 / 2,1}, then the value range of F (2)
The quadratic function f (x) = x & # 178; - (A-1) x + 5 is a monotone function in the interval (1 / 2,1),
It is shown that the axis of symmetry is outside the interval, that is, the axis of symmetry ≤ 1 / 2 or the axis of symmetry ≥ 1, and the union set can be obtained, so that the range of a can be solved
F (2) is the first-order expression of A. the range of a is obtained. If you deform f (2), you can get it
The minimum positive period sum of F (x) = cosx SiNx
f(x)=cosx-sinx
=(√2)cos(x+(π/4))
-√2≤f(x)≤√2
T=2π