It is known that M is the sum of all integers a satisfying the inequality root 3 < a < root 6, and N is the largest integer solution satisfying the inequality x ≤ 2 / 2 of the root sign 37-2 Square root of

It is known that M is the sum of all integers a satisfying the inequality root 3 < a < root 6, and N is the largest integer solution satisfying the inequality x ≤ 2 / 2 of the root sign 37-2 Square root of

2=√4＜√6＜√9=3
-2=-√4＜√3＜-√1=-1
M=-1+0+1+2=2
2=（√36-2）/2＜（√37-2）/2＜（√49-2）/2=2.5
Then n = 2
M+N=4

The integer x satisfying negative root 5 < x < root 3 is If it's right, you'll get 100 points

—√5＜x＜√3
Since √ 5 is between 2 and 3, so - √ 5 is between - 3 and - 2, and X takes the smallest integer - 2
Since √ 3 is between 1 and 2, X takes the largest integer as 1
Therefore, X takes integers between - 2 and 1, so x takes four integers - 2, - 1, 0 and 1

0

-1,0,1,2

The solution set of inequality [(x-1) * radical (x ^ 2-x-2)] > = 0?

0

0

Radical 2 * (x - radical 3) = Radix 6 * (x + 1)
Root 2 * x - root 6 = root 6 * x + root 6
Root 2x root 6x = 2 Radix 6
X = 2 root 6 / (root 2-root 6)
Multiply root 2 + root 6
X = [2 root 6 * (root 2 + root 6)] / 2-6
X = 2 root sign 12 + 12 / (- 4)
=-Radical 3-3

The solution set of Tana + radical 3 > 0 is

Tana + radical 3 > 0
tana＞√3
2kπ+π/3

1 + Tana / 1-tana = 3 + 2 times root number 2 Tana =?

1 + Tana / 1-tana = 3 + 2 times root number 2
Multiply cosa up and down at the same time
Cosa + Sina / cosa Sina = 3 + 2 times root number 2
Square the two sides
1 + sin2a / 1-sin2a = (3 + 2 times root 2) ^ 2
Then the solution is obtained
Sin2a = (16 + 12 times root number 2) / (18 + 12 times root number 2)
Sin2a
=2 (4 + 3 times root number 2) / 3 (3 + 2 times root number 2)
=2 root number 2 (3 + 2 times root number 2) / 3 (3 + 2 times root number 2)
=2 root sign 2 / 3
be
Sinacosa = 1 / 2sin2a = radical 2 / 3
If 1 + Tana / 1-tana = 3 + 2 times root number 2, then Tana > 0
So cosa and Sina are the same
sinA+cosA
And (Sina + COSA) ^ 2 = 1 + sin2a = (3 + 2 times root 2) / 3
So Sina + cosa = (root 6 + root 3) / 3, - (root 6 + root 3) / 3

In the same plane rectangular coordinate system, the images of functions f (x) = radical X and G (x) = X-1 are drawn, and the solution set of inequality root x greater than X-1 is obtained by using the images

Can an image be drawn? The first is a power function, and the second is a first-order function
The root x > X-1 means that the image of root x is above the X-1 image

The solution set of inequality radical 3 < 2x is

x>√3/2

Solving inequality (X-2) * root sign (x ^ 2-2x-3) ≥ 0 The answer is {x = {x}

Inequality (X-2) * √ (x ^ 2-2x-3) ≥ 0
1. First of all, in order to make the things under the root sign meaningful, x ^ 2-2x-3 ≥ 0
Thus, X ≥ 3 or x0 and √ (x ^ 2-2x-3) > 0 are obtained
（2） X-23
Solution (3)
X-2 = 0, x = 2
Solution (4)
If (x ^ 2-2x-3) = 0, x = - 1 or x = 3
Therefore, the final result is x ≥ 3 or x = - 1