When k is a value, the equation (K + 1) cosx + 4cosx-4 (k-1) = 0 has a real solution

When k is a value, the equation (K + 1) cosx + 4cosx-4 (k-1) = 0 has a real solution

When k is a value, the equation (K + 1) cos * x + 4cosx-4 (k-1) = 0 has a real solution, and the equation becomes k = (2 + cosx) / (2-cosx), then the solution of X is equivalent to the value range of K. geometrically, K is the slope of the line from point m (2,2) to line y = - x (x ∈ [- 1,1])
If x1x2 is the root of the equation x square-3x-2 = 0, then the value of X1 + X2 is? Is there any other way to solve the root formula
Direct use of Veda's theorem
x1+x2=3
It is known that the quadratic equation x + (4m + 1) x + 2m-1 = 0
1. Verification: no matter m is any real number, there are always two unequal roots of the equation. 2. If the two roots of the equation are X1 and X2, and satisfy the condition of 1 ﹣ X1 + 1 ﹣ x2 = - (1 ﹣ 2), find the value of M
(1) B - 4ac = (4m + 1) - 4 (2m-1) = 16m + 8m + 1-8m + 4 = 16m + 5, because 16m + 5 ＞ 0, the equation has two unequal real roots. (2) X1 + x2 = - (4m + 1), X1 * x2 = 2m-1; 1 △ X1 + 1 △ x2 = (x1 + x2) / X1 * x2 = - 1 / 2, X1 * x2 = - 2 (x1 + x2), that is 2m-1 = - 2 [- (4m + 1)] 2m -
1 a = 1 b = 4m + 1 C = 2m-1 △ = b = 4ac = 16m + 8m + 1-8m + 4 = 16m + 5, so △ 0, so the equation always has two unequal real roots. The solution is: M = - 1 / 2
If X: Y: z = 2:3:4, find the value of X + y + Z / x-2y + 3Z
x:y:z=2:3:4
Let x = 2K, y = 3k, z = 4K
(x+y+z)/(x-2y+3z)
=(2k+3k+4k)/(2k-2*3k+3*4k)
=9k/(8k)
=9/8
Let X: Y: z = 2:3:4 = k, then x = 2K, y = 3k, z = 4K, so x + y + Z = 9K, x-2y + 3Z = 8K, so the original formula is 9 / 8
There's no solution
Let x = 2A
x:y:z=2:3:4
y=3a
z=4a
(x+y+z)/(x-2y+3z)=(2a+3a+4a)/(2a-6a+12a)=9a/8a=9/8
=2x+3x+4x+4/2+2（3x）+3（4x）
=9x+2-6x+12x
=15x+2
I think so. Well, good luck
If the equation cos * 2x + 4sinx-a = 0 has a real number solution in the interval (0, π / 2), the value range of real number a is obtained?
(cosx)^2+4sinx-a=0
1-(sinx)^2+4sinx=a
-(sinx-2)^2+5=a
When 0
The roots of equation x are X1 =, X2 =, X1 + x2 =, x1x2 =;
X1 = [- B + radical (b ^ 2-4ac)] / 2A
X2 = [- b-radical (b ^ 2-4ac)] / 2A
X1+X2=-(b/a)
X1X2=c/a
MX ^ 2-4x ^ 2 = x + 1 is a quadratic equation of one variable about X, then the value range of M is?
The results show that: (M-4) x ^ 2-x-1 = 0
Because the equation is quadratic,
So M-4 ≠ 0
m≠4
Given (| x + 1 | + | X-2 |) (| Y-2 | + | y + 1 |) (| Z-3 | + | Z + 1 |) = 36, find the maximum and minimum of X + 2Y + 3Z
Graphic method = = please upload pictures = = if I understand and think it is good, I will offer a reward (how much confidential = = but there must be some)
It seems that few people can upload pictures = = so lower requirements = = don't be someone else's answer, give me a clear explanation = = I searched on Baidu and didn't understand. So come here to ask questions ~ please don't plagiarize~
Let me talk about the meaning of this formula first. |x + 1 | + | X-2 | is equivalent to | X - (- 1) | + | X-2 | is the sum of the distances from a point on the X axis to - 1 and 2. If you put X between - 1 and 2, obviously the shortest distance is 3. If you put X in other places, there will be repetition, the distance will be greater than 3
The equation cos ^ 2-sin ^ 2 + SiNx = m + 1 has a real number solution to find the value range of real number M
cos²x-sin²x+sinx=m+11-2sin²x+sinx=m+1m=-2sin²x+sinx=-2(sinx-1/4)²+1/8-1≤sinx≤1 -5/4≤sinx-1/4≤3/4 0≤(sinx-1/4)²≤25/16 -25/8≤-2(sinx-1/4)²≤0-25/8+1/8≤-2(sinx-...
Ah
The solution is to replace cos ^ 2 with 1-sin ^ 2, and then find the range on the right side of the equation. The result m belongs to the closed interval 9 / 8,43 / 16
Let x 2 + x 2 be the roots of two absolute equations
From ln X-2 = m
e^m=｜x-2｜
x-2=±e^m
x1=2+e^m x2=2-e^m
X1 + x2 = 4 (^ followed by exponent, for example, 3 ^ 5 means the 5th power of 3)