If the positive integer AB is the two roots of the equation x ^ 2 - ((a ^ 2-9) / 13) x + 10B + 56 + 5 √ (205 + 52b) = 0 about X, find the value of ab

If the positive integer AB is the two roots of the equation x ^ 2 - ((a ^ 2-9) / 13) x + 10B + 56 + 5 √ (205 + 52b) = 0 about X, find the value of ab

-(a^2-9)/13=a+b
10b+56+5√(205+52b)=ab

Given that Tana = - 1 / 3, tanb = 2, and a, B ∈ (0, π), then a + B

tan(a+b)
=(tana+tanb)/(1-tanatanb)
=(-1/3+2)/(1+1/3*2)
=(5/3)/(5/3)
=1
Because a, B ∈ (0, π)
And because of tana0
b∈(0,π/2）
a+b∈(π/2,3π/2)
So a + B = 5 π / 4

Given the point P (3 + m, 2n) and Q (2m-3, 2n + 1), and PQ ‖ Y axis, then the values of M and N are

PQ ‖ Y axis indicates that the abscissa of P and Q points are equal, that is to say
3+m=2m-3;
m=6;
The ordinate can't be determined, so n can't be determined

Given the function f (x) = sin ^ 2x + root 3sinxsin (x + π / 2) - A, (a ∈ R, a is a constant)
Finding monotone increasing interval and minimum positive period of function f (x)
If the absolute value of the inequality f (x) is less than 2, it holds on X ∈ [π / 4, π / 2]
When the function f (x) takes the maximum value, find the set of independent variables X
Online, etc

In this paper, we use the double angle formula to simplify sin (2x) and COS (2x), and then combine them, and then seek the derivative to find the monotone increasing or decreasing interval. Take the reciprocal zero, find the extreme value, and then find the maximum value

The linear equation of intersection of two straight lines 3xy-5 = 0 and 2x-3y 4 = 0 with equal intercept on two coordinate axes is obtained

3X+Y-5=0 (1)
2X-3y+4=0 (2)
y=5-3x
2x-3(5-3x)+4=0
2x-15+9x+4=0
x=1
y=5-3=2
So the intersection (1,2)
Let the intercept be a
Then the line is x / A + Y / a = 1
x+y=a
a=1+2=3
So x + y + 3 = 0

A simple method is used to calculate 1.25 * 42 - (4.46 + 0.14) =?

1.25*42-（4.46+0.14）
=1.25x40+1.25x2-4.6
=1.25x8x5+2.5-4.6
=50+2-4.6
=52-4.6
=47.4

It is known that O is the origin of the coordinate, and the line L passing through point P (2,1) intersects with the positive half axis of X axis and Y axis at two points a and B respectively
If line L and line y = 2x intersect at point C, the minimum area of triangle AOC and the equation of line L are obtained

Let the slope of l be K, then K 2x-1 = K (X-2) = > x = (2k-1) / (K-2) y = (4k-2) / (K-2) ‖ the area of AOC of triangle s = 1 / 2 | OA | * | 4k-2 | / | K-2 | = (2-1 / K) * (2k-1) / (K-2) = (2k-1) / (K-2) = (2k-1) / # 178; / (K & # 178; - 2K) = (4K & # 17

Let the sum of the first n terms of the sequence {an} be SN. For all n ∈ n * points (n, Sn), find the expression of an on the image of the function f (x) = x ^ 2 + X

Because the point (n, Sn) is on the image of function f (x) = x ^ 2 + X, then Sn = n & sup2; + N, then sn-1 = (n-1) & sup2; + (n-1), so sn-sn-1 = an = 2n

(1 / 3) there is a point charge with q = minus 3 times 10 to the minus 6 power C, which moves from point a to point B in an electric field, and the charge overcomes the point field force to do 6 times 10 to the minus 4
(1 / 3) there is a point charge with q = negative 3 times 10 to the negative sixth power C, which moves from point a to point B in an electric field. The charge overcomes the point field force to do the work of 6 times 10 to the negative fourth power J, and moves from point B to point C,

WAB=q*UAB
6*10^(-4)=3*10^(-6)*UAB
UAB = 200V
WBC=q*UBC
-9*10^(-4)=3*10^(-6)*UBC
UBC = - 300 V
UCA = UCB + UBA = - (UBC + UAB) = - (- 300 + 200) = 100V
The potential at point C is 300V,
The potential at point C is higher than that at point a

The inequality x2-ax + 1 ≥ 0 holds for all a ∈ [1,2], then the value range of real number X

Let X & # 178; - ax + 1 be regarded as a function of degree f (a) = - XA + (X & # 178; + 1)
So long as:
F (1) ≥ 0 and f (2) ≥ 0
The results are as follows
X & # 178; - x + 1 ≥ 0 and X & # 178; - 2x + 1 ≥ 0
X can go to all real numbers