Find the integer solution of equation 1 / x + 1 / Y-1 / [x (the square of Y)] = 3 / 4

Find the integer solution of equation 1 / x + 1 / Y-1 / [x (the square of Y)] = 3 / 4

Take x as a constant, use the denominator and the root formula to find y
－4x+/_ 4 radical (x * x-3x + 4)
y= -----------------------
2(4-3x)
If y is an integer, then the root sign (x * x-3x + 4) must be an integer
So if x = 3, then y = 2

We know the equation of curve C: x ^ 2 + y ^ 2-4x + 2Y + 5m = 0, if M = 0, is there a straight line L passing through point P (0,2) intersecting with curve C at two points a and B, and | PA | = | ab |, if there is an equation of L, if not, explain the reason!

When m = 0, the curve is: X \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ifl: y = KX + 2, then:
D = | 2K + 3 | / √ (1 + K & # 178;) = √ 3, the solution is: K & # 178; + 6K + 6 = 0, the value of K is obtained, that is, the equation of line L is obtained

Given that circle x ^ 2 + y ^ 2-2x + 2y-3 = 0 and circle x ^ 2 + y ^ 2 + 4x-z = 0 are symmetric with respect to line L, the equation of line L is obtained

-Is Z - 1?
(x-1)^2+(y+1)^2=5
(x+2)^2+y^2=5
Center a (1, - 1), B (- 2,0)
Symmetry is the center of the circle
So l is the perpendicular of ab
The slope of AB is (0 + 1) / (- 2-1) = - 1 / 3
So the slope of the vertical is 3
And passing the midpoint of AB (- 1 / 2, - 1 / 2)
So it's 3x-y + 1 = 0

When a > 0, B = 0, C

You can't be sure
This is related to the value of AC
So I'm not sure
What do not understand can continue to ask, online at any time, etc

Find the maximum and minimum of cube-3x of F (x) = x on [negative radical 3, positive radical 3]

Cubic-3x of F (x) = x
F '(x) = the square of 3x-3 = 3 (the square of x-1) = 0
x1=-1,x2=1
x -3 （-3,-1） -1 （-1,1） 1 （1,3） 3
f‘（x） + 0 - 0 +
F (x) - 18 increase maximum 2 decrease minimum - 2 increase 18
So when x = - 3, we get the minimum value of - 18, and when x = 3, we get the maximum value of 18

Y = (x ^ 2 + 5) / under the root sign (x ^ 2 + 4), find the range. I want to know why we can't use the mean inequality to find the maximum value
Isn't (x ^ 2 + 4) under the root sign and (x ^ 2 + 4) under the root sign the same sign? Why can't we go to the equal sign?

y=（x^2+5)/√(x^2+4）
=[(x^2+4)+1]/√(x^2+4)
=√(x^2+4)+1/√(x^2+4)
Three conditions must be satisfied to find the maximum value by means inequality
All items are positive
2. Definite: the product of the summation is the definite value
The sum of the maximum value of the product is constant
The two terms (a, b) involved can be equal
1 & # 186; √ (x ^ 2 + 4) > 0,1 / √ (x ^ 2 + 4) > 0
2 & # 186; √ (x ^ 2 + 4) × 1 / √ (x ^ 2 + 4) = 1, consistent with
If √ (x ^ 2 + 4) = 1 / √ (x ^ 2 + 4)
Then x ^ 2 + 4 = 1 = = > x ^ 2 = - 3
∵ x ∈ R, x ^ 2 = - 3 does not hold
√ (x ^ 2 + 4) and 1 / √ (x ^ 2 + 4) cannot be equal
We can't use the mean inequality to get the maximum value
Let √ (x ^ 2 + 4) = t ≥ 2
Y = t + 1 / T, increasing on [2, + ∞)
When t = 2, the minimum value of Y is 5 / 2

Is there a root sign (a ^ 2 + B ^ 2) > = root sign (2 * a * b) in the mean inequality?

The inequality you asked should belong to a derivation of mean inequality
It comes from the mean inequality a ^ 2 + B ^ 2 > = 2Ab ①
① The so-called formula can be fixed as the mean inequality because no matter what value a or B takes, its inequality is always true
The inequality you asked is squared on both sides of Formula 1. When the signs of a and B are different, it is wrong to open the root sign at will
Therefore, this inequality is not only a mean inequality, but also a derivation under the condition that a and B have the same sign
In addition, write a little more, no matter what the circumstances, to the variable root, must take the absolute value on the result, this is often the test point and loss point

The use of mean inequality of 1-x Λ 2 under 1 / 2x times root sign

1-x & # 710; 2 + X & # 710; 2 = 0, so the former 1 / 2 does not move, and the latter uses inequality
x/2*√(1-xˆ2)

The maximum value of y = x (1-3x ^ 2) is determined by means inequality or Cauchy inequality

First of all, if you square both sides, you can get y ^ 2 = x ^ 2 (1-3x ^ 2) ^ 2
Then according to the mean inequality, x ^ 2 (1-3x ^ 2) ^ 2 = 1 / 6 * 6 * x ^ 2 (1-3x ^ 2) ^ 2 ≤ 1 / 6 * (2 / 3) ^ 3 = 4 / 81

X belongs to 0 to 1 / 3, use the basic inequality to find the maximum value of y = x * (1-3x) ^ 0.5

y=x√(1-3x)
=(2 / 3) √ (3x / 2) √ (3x / 2) √ (1-3x)
≤ (2 / 3) {[(3x / 2) + (3x / 2) + 1-3x] / 3} ^ (3 / 2) [using inequality]
=2√3/27
If and only if 3x / 2 = 1-3x, that is, x = 2 / 9, the equal sign holds
The inequality step uses: A & # 178; + B & # 178; + C & # 178; ≥ 3 [(ABC) &# 178;] ^ (1 / 3),
ABC ≤ [(A & # 178; + B & # 178; + C & # 178;) / 3] ^ (3 / 2)