If the maximum value of F (x) = 6x + B / X & # 178; + 4 is 9 / 4, then B is a real number___

If the maximum value of F (x) = 6x + B / X & # 178; + 4 is 9 / 4, then B is a real number___

y=(6x+b)/(x^2+4)=9/4(6x+b)
x^2-27x/2+4-9b/4>=0
If we want to make it constant greater than or equal to 0 and get the equal sign, then it is a complete square
delta=(27/2)^2-4(4-9b/4)=729/4-16+9b=665/4+9b=0-->b=-665/36
b> 0, there is no maximum,
When B = 0, it is a straight line with no maximum
So B = - 3 radical (3x * 3x * B / x) = - 9 radical B = - 7 / 4
b=(7/36)²
Find the coefficient of x ^ 3 in the product of polynomial x ^ 3-2x + 1 and 3x ^ 2 + 5x-7
(x^3-2x+1)(3x^2+5x-7)
=3x^5+5X^4-7x^3-6X^3-10X^2+3x^2+5x-7
=3x^5+5X^4-13x^3-7x^2+5x-7
Coefficient with x ^ 3: - 13
If the system of inequalities {x 1} has no solution, find the range of A
2x-1 / 3 > 1 is (2x-1) / 3
(2x-1)/3>1
2x-1>3
2x>4
X>2
Because it's related to X
2x-1/3>1
X > 2 / 3
For X2, there are: a
If | a | = 1, find the solution of the univariate quadratic equation (A-1) x & sup2; + (a + 5) x + 2 = 0
∵a | = 1 ∵ a = 1 or a = - 1 ∵ is the univariate quadratic equation (A-1) x & sup2; + (a + 5) x + 2 = 0 ∵ A-1 ≠ 0 (the quadratic coefficient of the univariate quadratic equation is not 0) ∵ a ≠ 1, that is, the equation a = - 1 becomes - 2x & sup2; + 4x + 2 = 0x & sup2; - 2x-1 = 0x = 1 ± √ 2
|A | = 1, a = 1 or a = - 1
When a = 1, the original univariate quadratic equation (A-1) x & sup2; + (a + 5) x + 2 = 0 is reduced to 6x + 2 = 0, x = - 1 / 3
When a = - 1, the original univariate quadratic equation (A-1) x & sup2; + (a + 5) x + 2 = 0 is simplified to x ^ 2-2x-1 = 0, x = 1 ± √ 2
When - 2 ≤ x ≤ 1, the maximum value of quadratic function y = - (X-H) &# 178; + 8 is 4, then the value of real number h is---------
When h is in the interval [- 2,1], the maximum value is y (H) = 8;
When H > 1, the maximum value is y (1) = - (1-h) ^ 2 + 8 = 4, then H = - 1 or 3, take H = 3
When h
The coefficient of the x ^ 3 term in the product of polynomials (3x ^ 4-2x ^ 3 + x ^ 2-8x + 7) (2x ^ 3 + 5x ^ 2 + 6x-3) is
kuai
-2x^3*(-3)+x^2*6x-8x*5x^2+7*2x^3=-14x^3
So the coefficient of x ^ 3 is - 14
If x-3 (X-2) x has no solution, what is the range of a?
x-3x+6=1
(a+2x)/3>x
a+2x>3x
X
X-3(X-2)≤4
X≥1
(A+2X)/3>X
A>X
If we want to make the original inequality have no solution, we must break this condition
So a ≤ x
A≤1
Solving the quadratic equation of one variable about X: x ^ 2 - (m ^ 2 + 5) x
1. If real numbers m and N satisfy (m ^ 2 + n ^ 2) ^ 2-3 = 2 (m ^ 2 + n ^ 2), then m ^ 2 + n ^ 2 =?
2. Solve the univariate quadratic equation about X: x ^ 2 - (m ^ 2 + 5) x + 2 (m ^ 2 + 3) = 0
1. If the real numbers m and N satisfy (m ^ 2 + n ^ 2) ^ 2-3 = 2 (m ^ 2 + n ^ 2),
（m^2+n^2)^2-3-2(m^2+n^2)=0
(m^2+n^2-3)(m^2+n^2+1)=0
m^2+n^2>=0
m^2+n^2+1>=1
m^2+n^2=3
2. Solve the univariate quadratic equation about X: x ^ 2 - (m ^ 2 + 5) x + 2 (m ^ 2 + 3) = 0
x^2-(m^2+5)x+2(m^2+3)=(x-2)(x-m²-3)=0
x1=2
x2=m²+3
Given that real numbers x and y satisfy x ^ 2 + 3x + Y-3 = 0, the maximum value of X + y is the problem in the step of solving the problem
Solution steps:
Let x + y = K
The solution is y = K-X and substituted by x ^ 2 + 3x + Y-3 = 0
x^2+3x+k-x-3=0
x^2+2x+k-3=0
Let △ = 4-4 (K-3) = 0
The solution is k = 4
So the maximum value of X + y is 4
Why let △ = 4-4 (K-3) = 0?
Here is a brief step,
△=4-4(k-3)=16-4k
Because the equation x ^ 2 + 2x + K-3 = 0 has a real root, so △≥ 0, that is 16-4k ≥ 0
So K ≤ 4
So the maximum value of K is 4
It should be
Because X has meaning, so in order to have a solution
△=4-4(k-3)>=0
4-4k+12>=0
K
If the polynomials 3x & # 179; - KX & # 178; - 2x & # 179; - 2x & # 178; + X-1 combine the similar terms without quadratic terms, the value of K is obtained
So K + 2 = 0;
So k = - 2;
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