Find Lim [(1 + 5 / N) to the nth power] This is a high number

Find Lim [(1 + 5 / N) to the nth power] This is a high number

Let 1 / a = 5 / n
n=5a
Original formula = LIM (a →∞) (1 + 1 / a) ^ 5A
=lim(a→∞)[(1+1/a)^a]^5
The limit of (1 + 1 / a) ^ A is e
So the original formula = e ^ 5

Calculation: LIM (n →∞) (n + 1 power of 2-A) / (n-1 power of 3-A)=

1. The absolute value of a > 1, limit = 1
2. Absolute value of a = 1, limit = 1 / 2
3. Absolute value of a

It is known that the equation 2kx2-4x-3k = 0 about X has two real roots x1, X2, and X11. Try to find the value range of real number K

(!) let p be true and Q false
So: because the equation x to the second power + MX + 1 = 0 has two unequal negative roots, let two be x1, x2,
Therefore, X1 + x2 = - M < 0, so m > 0
Because the equation 4 (the second power of x) + 4 (m-2) x + 1 = 0 has no real root, it is false
M > 3 or m < 1
To sum up: 0 < m < 1
(2) P false, Q true:
The second power + MX + 1 = 0 of equation x has two unequal negative roots, so m < 0
The equation 4 (the second power of x) + 4 (m-2) x + 1 = 0 has no real root and is true, so:} ≤ 0, that is: 16 (m-2) ^ 2-16 ≤ 0,
1 ＜m＜3.
To sum up, when p is false and Q is true, M has no solution
So: 3 ≤ m or 1m ≤ 2

Let two of the equations 2x ^ 2 - (2k-2) x - (2 / 3K ^ 2 + 5 / 6) = 0 be X1 and X2 respectively, and | X1 / x2 | = 2 to find the value of K

Because | X1 / x2 | = 2
1. When X1 / x2 = 2
x1=2x2
From Veda's theorem
x1*x2=-1/3k^2-5/12=2x2^2……………… one
x1+x2=3x2=k-1……………………………… two
From 2
9x2^2=(k-1)^2
Because = - 1 / 3K ^ 2-5 / 12 = 2x2 ^ 2
So the two styles stand together
We get x2 = k=…………
I don't understand. I'll take it myself
2. When X1 / x2 = - 2
…………
Same thing
Find X2, K
I wish you study every day, come on!

Simple calculation of 18 * 0.86 + 1.8 * 1.3 + 0.18

18*0.86+1.8*1.3+0.18
=1.8*8.6+1.8*1.3+1.8×0.1
=1.8×（8.6+1.3+0.1）
=1.8×10
=18

In triangle ABC, if a ^ 3 + B ^ 3-C ^ 3 divided by a + B-C equals 0, and Sina × SINB = 3 / 4, judge the shape of triangle

According to the meaning a ^ 3 + B ^ 3 = C ^ 3, C is the largest edge
According to the sine theorem, sinc > Sina, sinc > SINB,
If a ^ 2 + B ^ 2

There is a natural number, which can be used to remove 226 odd a, 411 odd a + 1 and 527 odd a + 2, then a=______ ．

410-226 = 184 = 23 × 8525-226 = 299 = 23 × 13525-410 = 115 = 23 × 5 19, that is: a = 19; so the answer is: 19

Through the fixed point m (m, 0), (M > 0) on the symmetry axis of the parabola y2 = 2x, make a straight line AB intersecting the parabola at two points a and B. (1) try to prove that the product of the ordinates of two points a and B is a fixed value; (2) if the minimum area of △ OAB is 4, find the value of M

(1) Let a (x1, Y1), B (X2, Y2) let a (x1, Y1), B (X2, Y2) be a (x1, Y1) and let a (x1, Y1), B (X2, Y2) △ = 4T2 + 8m ＞ 0, Y1 + y2 = 2T, Y1 = 2T, y2 = 2T, y2 = 2T, y2 = 2m, the constant \\\\951; Y1 = 2 = the constant \\\\\\\\\\\\\\\\\\\\\\\\\\\\m = 4 {M = 2

If K is an integer and x2-kx-15 can decompose a factor, then K=_____ The factorization is -——

-15 = - 1 * 15 = 1 * (- 15) = - 3 * 5 = 3 * (- 5), so x ^ 2-kx-15 can be decomposed into:
1. (x-1) (x + 15), expand to: x ^ 2 + 14x-15, then k = - 14;
2. (x + 1) (X-15), expand to: x ^ 2-14x-15, then k = 14;
3. (x-3) (x + 5), expand to: x ^ 2 + 2x-15, then k = - 2;
4. (x + 3) (X-5), expand to: x ^ 2-2x-15, where k = 2;
In conclusion, k = ± 2, ± 14
Factorization:
x^2-kx-15=(x-1)(x+15)、(x+1)(x-15)、(x-3)(x+5)、(x+3)(x-5)

Given the non right triangle △ ABC, the angle a = 45 ° and the intersection of the straight line of high BD and CE at point O, the degree of the angle BOC is calculated

1. C is the acute angle
∠ABD=45,∠BoC=45+90=135
2. C is an obtuse angle
The intersection of O and triangle is 45 degrees