Mathematics: 2 / (n) + 3 / (n + 1) + 4 / (n + 2) = 133 / 60 (n is the denominator), find the positive integer of N?

2/(n)+3/(n+1)+4/(n+2)=133/60
No matter what the form of the numerator is, the denominator must be n (n + 1) (n + 2), which is the product of three consecutive integers. If you want to make it equal to the right, then n (n + 1) (n + 2) must be a multiple of 60. Considering that 60 can be decomposed into 3 × 4 × 5, so n can only be equal to 3!

Try to determine, for which positive integer a, equation 5x2-4 (a + 3) x + a2-29 = 0 has positive integer solution? All positive integer solutions of the equation are obtained

The equation is rewritten as & nbsp; (X-6) 2 + (a-2x) 2 = 65. Since 65 is expressed as the sum of the squares of two positive integers, there are only two different forms: 65 = 12 + 82 = 42 + 72, so | & nbsp; X & nbsp; - & nbsp; & nbsp; 6 & nbsp; | = 8 | a-2x | = 1 ① , or & nbsp; | & nbsp; X & nbsp; - & nbsp; & nbsp; 6 & nbsp; | = 7 | - 2x | = 4 ②| x - 6 |=1|a-2x|=8… ③ , or & nbsp; | & nbsp; X & nbsp; - & nbsp; & nbsp; 6 & nbsp; | = 4 | - 2x | = 7 ④ From ① we get x = 14 (when a = 29 or 27); from ② we get x = 13 (when a = 22 or 30); from ③ we get x = 5 (when a = 2 or 18); & nbsp; or & nbsp; X = 7 (when a = 6 or 22); from ④ we get x = 2 (when a = 11); or & nbsp; X = 10 (when a = 13 or 27)

Try to determine for what kind of integer a, the equation 5x-4 (a + 3) x + a-29 = 0 has integer solutions? And find all positive integer solutions of the equation

A ^ 2-4xa + 4x ^ 2 + x ^ 2-12x-29 = 0 (a-2x) ^ 2 + (X-6) ^ 2 = 65 a-2x and X-6 are integers, and 65 can only be decomposed into 1 ^ 2 + 8 ^ 2 4 ^ 2 + 7 ^ 2 X-6 = 8 x = 14 a-2x = 1 a = 29 a-2x = - 1 a = 27 X-6 = 1 x = 7 a-2x = 8 A = 22 a-2x = - 8 A = 6 X-6 = - 1 x = 5 a-2x = 8 A = 18 a

Let the equation of the straight line be (a + 1) x + y + 2 + a = 0 (a belongs to R). If the intercept of the straight line L on the two coordinate axes is equal, the equation of the straight line L is obtained

When the intercept is equal to 0, 2 + a = 0, a = - 2, the linear equation

Simple calculation method: 3.4 times 7.5 + 34 times 0.25

3.4 times 7.5 + 34 times 0.25 = 34 * 0.75 + 34 times 0.25 = 30 * (0.75 + 0.25) = 34 * 1 = 34

If the line y = KX + B is parallel to the line y = (2-x) / 3 and intersects with the line 2x + 3Y + 1 = 0 at the same point on the Y axis, then K= ,b__ .

y=(2-x)/3=-x/3+2/3
Y = KX + B is parallel to the line y = (2-x) / 3
So k = - 1 / 3
y=-x/3+b
Where is the intersection of the line 2x + 3Y + 1 = 0 and the Y axis
x=0,y=-1/3
(0,-1/3)
So (0, - 1 / 3) is on y = - X / 3 + B
-1/3=0+b
So k = - 1 / 3, B = - 1 / 3

The sum of the first n terms of the proportional sequence an is SN. It is known that for any n, the positive integer points (n, Sn) are in the image of the function y = B ^ x + R (b > 0 and B ≠ 1, B, R are constant
Find R

The points (n, Sn) are all on the image of the function y = B ^ x + R,
Sn=b^n+r,
When n = 1, A1 = S1 = B + R
When n ≥ 2, an = SN-S (n-1) = B ^ N-B ^ (n-1) = (B-1) B ^ (n-1)
If {an} is an equal ratio sequence, then A1 = B + R should conform to an = (B-1) B ^ (n-1),
So B + r = (B-1) B ^ 0, B + r = B-1
r=-1.

There are two point charges in vacuum, Q1 = + 3 times the negative octave of 10 and Q2 = - 3 times the negative octave of 10. The distance between them is 0.1M
The distance between point a and two point charges is 0.1M
Work it out for me, thank you

The field strength of point charge 1 at point a is e (1) = KQ / r2 = 2.7 times 10
The field strength of point charge 2 at point a is also e (2) = KQ / R 2 = 2.7 times 10
Because the distance between the two point charges and point a is equal, it is an equilateral triangle. E (1) and E (2) form their external angle of 120 degrees. The sum of their vectors is the third power of E = 2.7 times 10

If the equation | 1-x | = MX has a solution, then the value range of real number M______ ．

|1-x | = MX, ① when x ≥ 1, X-1 = MX, (1-m) x = 1, m ≠ 1, x = 11 − m, ｜ 11 − m ≥ 1, the solution is 0 ＜ m ＜ 1; ② when x < 1, 1-x = MX, (1 + m) x = 1, m ≠ - 1, x = 11 + m, 11 + m ＜ 1, ｜ 1 + m ＜ 0 or 1 + m ≥ 1, ｜ m ＜ - 1 or m ≥ 0; in conclusion: the solution set is: m ≥ 0

Given that a and 2b are reciprocal to each other, - C and D / 2 are opposite to each other | x | = 4, find 4ab-2c + D + X / 4

2ab=1
-c+d/2=0
x=±4
4ab-2c+d+x/4
=2x1-2x0+±4/4
=2±1
=3 or 1

Mathematics: 2 / (n) + 3 / (n + 1) + 4 / (n + 2) = 133 / 60 (n is the denominator), find the positive integer of N?