Is to determine the number of real roots of the equation x plus the absolute value lgx equal to 2

Is to determine the number of real roots of the equation x plus the absolute value lgx equal to 2

Equation x + | lgx | = 2, that is | lgx | = 2-x
Draw the image of y = lgx, and then set 0

Finding the number of real number solutions of the equation lgx = | cosx |

If you need to draw an image, as long as you draw an image, you can easily see the intersection of lgx and | cosx | image. If there are several intersection points, there are several real solutions. It should be three

If the equation cos2x + cosx-a = 0 has a solution, then the range of real number a is

From cos2x + cosx-a = 0 = = > 2cos & sup2; X + cosx - (a + 1) = 0
The equation cos2x + cosx-a = 0 has a solution,
Then a ≥ - 9 / 8
So if the equation cos2x + cosx-a = 0 has a solution, then the range of real number a is a ≥ - 9 / 8

Solution set of equation cos2x = cosx in [0,2 π]

From cos2x = cosx, we can get that:
2(cosx)^2 -1=cosx
SO 2 (cosx) ^ 2 - cosx - 1 = 0
That is: (2cosx + 1) (cosx - 1) = 0
The solution is: cosx = - 1 / 2 or 1
So in [0,2 π]
When cosx = - 1 / 2, x = 2 π / 3 or 4 π / 3
When cosx = 1, x = 0 or x = 2 π
The solution set of X in [0,2 π] is obtained
x∈{0,2π/3,4π/3,2π}

The equation 2x ^ 3 + 3x ^ 2 = 6x = 0 has several real roots
Be more specific. Thank you~

Analysis
simple form
2x3+3x²+6x=0
x(2x²+3x+6)=0
b²-4ac=9-4*2*6

The number of real roots of equation 3x ^ 2 + 6x-1 / x = 0 is
Process, thank you
Why

The most general method,
The original formula is the same as X
3x^3+6x^2-1=0
3x*(x^2+2x+1)-(3x+1)=0
Have (x + 1) ^ 2 = 1 + (1 / 3x)
By drawing their images (it should not be difficult? (x + 1) ^ 2 is y = x ^ 2, one unit to the left; 1 + (1 / 3x) is the function, y = 1 / 3x, one unit to the up), we can observe that there are three intersections, that is, the original equation has three solutions (I draw two in the second quadrant and one in the first quadrant)

The number of real roots of equation 3x ^ 2 + 6x = 1 / x?
Such as the title

It can be judged by drawing
Draw f (x) = 3x & sup2; + 6x respectively
And G (x) = 1 / X
It's obvious that there's one intersection in the first quadrant and two intersections in the third quadrant
So this equation has three real roots

If the cubic power of equation x - 3x + M = 0 has three different roots, then the range of real number m?
It's better to use chart method

Find the derivative and know that the value range of the three roots is (- infinity, - 1), (- 1,1), (1, infinity). Then only f (- 1) * f (1)

If the equation (A-1) x + 2 (a + 1) x + A + 5 = 0 has two real roots, find the value of positive integer a
I don't have any more points. Please do it for me
There will be good news
Before tonight
.

Because the equation (A-1) x + 2 (a + 1) x + A + 5 = 0 has two real roots
So A-1 ≠ 0
[2(a+1)]^2-4(a-1)(a+5)≥0
The solution is a ≠ 1
a≤3
So the value of positive integer a is 2,3

If the equation (a-b) x & sup2; - 8x + 6 = 0 of X has real roots, then the maximum value of integer is

Let me tell you, the title is misprinted, and the correct one should be: about the X equation (a-6) * x ^ 2-8x + 6 = 0 has real root, what is the maximum value of integer a? So you should know. The equation has real root, that is (- 8) ^ 2-4 (a-6) × 6 ≥ 064-24a + 144 ≥ 0-24a ≥ - 208a ≤ 26 / 3. The maximum value of a is 26 / 3