5x6 + 48 △ x = 36 to solve the equation

5x6 + 48 △ x = 36 to solve the equation

5x6+48÷X=36
30+48÷x=36
48÷x=36-30
48÷x=6
x=48÷6
x=8
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48-x + 3x7 = 36

48-X+3x7=36
48-x+21=36
-x=36-48-21
-x=-33
x=33

X1x2x3 is the three roots of the equation X1 ^ 3 + PX + q = 0, then the determinant (first row) X1 x2 X3 (second row) X3 X1 X2 (third row) x2 X3 x1

According to Weida's theorem, we get: X1 + x2 + X3 = 0,
First line X1 x2 x3
The second line X3 X1 x2
The third line x2 X3 x1
Add lines 2 and 3 to line 1, and the three numbers in the first line are X1 + x2 + X3
That is, the first line is 0
So the value of the original determinant is 0

If Tana = √ 3 (1 + a), √ 3 (Tana * tanb + a) + tanb = 0, a and B belong to (0,90 degrees), then a + B=

Tana = root 3 (1 + a) root 3 (Tana * tanb + a) + tanb = 0 so: 3 (1 + a) tanb + root 3A + tanb = 0 (4 + 3a) tanb = - root 3A so: tanb = - root 3A / (4 + 3a) because Tana = root 3 (1 + a) and Tan (a + b) = (Tana + tanb) / (1-tana * tanb) Tan (a + b) = root 3A + B

If the distance from point P to the x-axis is three unit lengths, and the distance to the y-axis is four unit lengths, then the coordinates of point P?

The distance from point P to the x-axis is three unit lengths, y = ± 3
The distance to the y-axis is 4 unit lengths, x = ± 4
Coordinates of point P: P1 (3,4) P2 (3, - 4) P3 (- 3,4) P4 (- 3, - 4)

Given that α > 0, the function f (x) = - 2 α sin (2x + π / 6) + 2 α + B, when x ∈ (0, π / 2), - 5 ≤ f (x) ≤ 1, ① find the values of constants A and B
② Let g (x) = f (x + π / 2) and LGG (x) > 0, find the monotone interval of G (x)

1 ,（0,π/2） 2x+π/6~(π/6,7π/6),sin(2x+π/6)~(-1/2,1],
From a > 0, so f (x) ~ [- 2A + 2A + B, a + 2A + b) = [- 5,1), so a = 2, B = - 5
2, f (x) = - 4sin (2x + π / 6) - 1 into G (x) reduction
If G (x) = 4sin (2x + π / 6) - 1, LGG (x) "0, then G (x) > 1, sin (2x + π / 6) > 1 / 2
2x+π/6~(2kπ+π/6,2kπ+5π/6),(kπ,kπ+π/3).
Corresponding increment: 2x + π / 6 ~ (2k π + π / 6,2k π + π / 2], (K π, K π + π / 6]
Decreasing: (K π + π / 6, K π + π / 3)
Pay attention to the opening and closing of the interval

Who knows that in the equation of straight line, a of intercept type is equal to 0, why is it parallel to X axis?

When the intercept a of a line on the x-axis is 0, the intercept is 0, which means that the line intersects through the x-axis and y-axis. It is not applicable to the linear equation x / A + Y / b = 1 in the form of intercept, but only applicable to the oblique intercept y = KX
A line parallel to the x-axis does not intersect with the x-axis, so its intercept on the x-axis does not exist (that is, there is no a), so the intercept formula can not be used
The main feature of this kind of line is that the ordinates of the points above it are the same: y = B. This is the equation of a line parallel to the X axis

Simple method calculation (process) - 0.25 ^ 14 * 2 ^ 30
A simple method for calculation
-0.25^14 * 2^30

-0.25^14 * 2^30
=-0.25^14×4^15
=-0.25^14×4^14×4
=-(0.25×4)^14×4
=-4

Given that the line L passes through the point P (2,1) and intersects with the x-axis and y-axis at two points a and B respectively, and O is the origin of the coordinate, the equation of the line L is obtained when the area of △ ABO is 4

y = - 1/2x + 2

It is known that SN is the sum of the first n terms of sequence {an}, and the points (n, Sn / N) (n ∈ n *) are all on the image of function y = 3x-2. The general formula of {an} is obtained
How to use s (n + 1) minus Sn, I always calculate A1 = 1 and an = 6N + 1

(I) according to the meaning of the title, that is, Sn =, when n ≥ 2, an = sn-sn-1 =;
When n = 1, A1 = S1 = 3 × 12-2 = 6 × 1-5, so an = 6n-5 (n ∈ n *)