Given that the equation 3x ^ 2-kx = 2-x has a root of - 2, what is the value of K?

Given that the equation 3x ^ 2-kx = 2-x has a root of - 2, what is the value of K?

We can get k by substituting the root - 2
3×4+2k=2+2
2k=-8
k=-4

What is the length and width of the inscribed rectangle of the ellipse x ^ 2 + 4Y ^ 2 = 4 and the largest area? (parameter)
What is the length, width and area of the inscribed rectangle of ellipse x ^ 2 + 4Y ^ 2 = 4?
(with parametric equation)
On April 3, please add + 5

Let the fixed point of the first quadrant be a (2cosa, Sina)
Then the length of the rectangle is 4cosa
Width of rectangle = 2sina
Area = 4sin2a
Take the maximum value when sin2a = 1
A = 45 degrees
Length of rectangle = 4cosa = 2 2
Width of rectangle = 2sina = root 2

2.3.4.5. In the order of five numbers, add the operation symbol, so that the calculated number is 10. What are the equations

(-1+2-3+4)*5=10
1*(2*3-4)*5=10
(1*2*3-4)*5=10
-1*2+3+4+5=10
1*[(2^3)/4]*5=10
(1^2-3+4)*5=10
1+2+3*4-5=10

Known function f (x) = (1-x) e ^ X-1
Known function f (x) = (1-x) e ^ X-1
If x ≥ 0, G (x) = e ^ x + λ ln (1-x) - 1 ≤ 0, calculate the value range of λ;
Prove: e ^ 1 / (n + 1) + e ^ 1 / (n + 2) + e ^ 1 / (n + 3) +. E ^ 1 / 2n < n + LN2 (n ∈ n *)

I only got the first question
Good trouble. I don't know if there is a better way

Ellipse and hyperbola have common focus, ellipse 25X ^ 2 + 9y ^ 2 = 1, the sum of their eccentricity is 2
It is known that hyperbola and ellipse 25X ^ 2 + 9y ^ 2 = 1 have common focus, and the sum of eccentricity of ellipse and hyperbola is 2

X & sup2; / (1 / 25) + Y & sup2; / (1 / 9) = 1 so a '& sup2; = 1 / 9, B' & sup2; = 1 / 25C '& sup2; = 1 / 9-1 / 25 = 16 / 225C' = 4 / 15 so e '= C / a = (4 / 25) / (1 / 3) = 12 / 25 so hyperbola e e = 2-e' = 38 / 25C = 4 / 15 focus on Y axis e = C / b = 28 / 25B = 5 / 21a & sup2; = C & sup2; - B & sup2

What is the meaning of ratio and reduction ratio?

The quotient obtained by dividing the preceding term of a ratio by the following term is called ratio
Reduction ratio is the simplest integer ratio

Find the tangent equation which passes through the point m (0, - 2) and is tangent to the curve f (x) = 2x ^ 3-x + 2
No, it's - 2

F (x) = 2x ^ 3-x + 2F ′ (x) = 6x ^ 2-1 let the tangent point be (m, n) ｜ the tangent slope be k = f ′ (m) = 6m ^ 2-1 ∵ the tangent passes through the tangent point (m, n), n = f (m) = 2m ^ 3-m + 2 ｜ the tangent equation is y-n = (6m ^ 2-1) (x-m) y - (2m ^ 3-m + 2) = (6m ^ 2-1) (x-m) ① ∵ the tangent passes through the point m (0, - 2) ∵ substituting the point m into the tangent

Given that L1: (3 + m) x + 4Y = 5-3m, l2:2x + (5 + m) y = 8, find out why m is the intersection of L1 and L2
〔5-3M -(3+M)X〕/4=(8-2x)/(5+m)
I don't know this step
How can we get K1 = - (3 + m) / 4, K2 = - 2 / (5 + m)

The intersection of lines means that the slopes are not equal or that one slope does not exist and the other slope exists
l1：（3+m)x+4y=5-3m,l2：2x+(5+m)y=8
So K1 = - (3 + m) / 4, K2 = - 2 / (5 + m)
If equal
-(3+m)/4=-2/(5+m)
(m+3)(m+5)=8
m^2+8m+7=0
m=-1,m=-7
So the intersection is m ≠ - 1, m ≠ - 7
If the slope does not exist, it is perpendicular to the X axis, that is, the coefficient of Y is 0
It's impossible
L2, M = - 5, m ≠ - 1, m ≠ - 7
So m ≠ - 1, m ≠ - 7

The formula for finding the last term in the arithmetic sequence is just a formula! No letter!

Last item = first item + (number of items - 1) * tolerance

(- 0.875-7 / 24 + 7 / 4) * - 8 / 7)

-0.875＝-7/8
（-0.875-7/24+7/4）*（-8/7）
＝（-7/8)*(-8/7) - (7/24)*(-8/7) + (7/4)*(-8/7)
=1+1/3 - 2
=-2/3