On the inequality MX & # 178; - 6mx + m + 8 ≥ 0 of X, the value range of M is obtained

On the inequality MX & # 178; - 6mx + m + 8 ≥ 0 of X, the value range of M is obtained

If M = 0, the constant holds
If it is not equal to 0, obviously, the opening must be upward, that is, M > 0
If the minimum value point is x = 3 and the minimum value > = 0 holds, then MX & # 178; - 6mx + m + 8 ≥ 0 holds
M (x ^ 2-6x + 1 + 8 / M) = m ((x-3) ^ 2-8 + 8 / M) > = 0, x = 3, then - 8 + 8 / M > = 0, M is obtained
A kind of
≥0
M
≥7
1. M = 0 MX2 + 6mx + m + 8 = 8 > = 0 holds
2. m≠0
M>0
Discriminant = 36m ^ 2-4m (M + 8)
=32m^2-32m
First simplify, then evaluate: xy-2x (Y-X) - 3x (x + y), where x = - 2, y = - 3
∵x=-2,y=-3
xy-2x(y-x)-3x(x+y),
=x(y-2y+2x-3x-3y)
=-x(4y+x)
=2(4×(-3)-2)
=-28
If the equation (m-1) x2-2mx-3 = 0 is a quadratic equation of one variable with respect to x, then the value range of M is______ .
Because the equation is a quadratic equation of one variable, so M-1 ≠ 0  m ≠ 1. So the answer is: m ≠ 1
The whole process of 2Y ^ 2-y-2 = - 1 matching method
solution
2y²-y-2=-1
2y²-y=1
y²-1/2y=1/2
(y²-1/2y+1/16)=1/2+1/16
(y-1/4)²=9/16
Ψ Y-1 / 4 = 3 / 4 or Y-1 / 4 = - 3 / 4
Or y = - 1
For - 1 ≤ a ≤ 1, the value range of X is ()
A. 0 ＜ x ＜ 2B. X ＜ 0 or X ＞ 2C. - 1 ＜ x ＜ 1D. X ＜ 1 or X ＞ 3
Let f (a) = x2 + (A-2) x + 1-A = (x-1) a + x2-2x + 1, ∵ - 1 ≤ a ≤ 1, the inequality x2 + (A-2) x + 1-A ＞ 0 hold, that is, X2 − 3x + 2 ＞ 0x2 − x ＞ 0, and the solution is: X ＜ 0 or X ＞ 2
It is known that the two roots of the equation x + (M-17) x + (m-2) = 0 about X are positive real numbers
The two roots are a, B: a + B = 17-m > 0 a * b = m-2 > 0 2
Is the equation (m-1) x & sup2; + 2mx + (M + 3) = 0 a quadratic equation with one variable? Try to find out the solution of the equation in different cases of m value
The formula x = (- B ± √ (b ^ 2-4ac)) / 2a is still used to get the expression of M
Discuss its > = 0 and
When m = 1, it is a linear equation with one variable, x = - 2
When m ≠ 1, it is a quadratic equation with one variable
M-1 is not equal to 0
The square of 4m - 4 * (m-1) (M + 3) is greater than 0
Then M is less than 3 / 2 and M is not equal to 1
X + 1 = 2Y, 2 / 2 x-3 / 3 y = 0
x+1=2y.1
x/ 2-y/3=0 .2
Formula 2
3x-2y=0.3
Substitute Formula 1 into formula 3
3x-(x+1)=0
3x-x-1=0
2x=1
x=1/2
x+1=2y
1/2+1=2y
2y=3/2
y=3/4
So x = 1 / 2, y = 3 / 4
When x ∈ (0,1), the inequality X & # 178;
X & # 178; & lt; loga (x + 1) is equivalent to loga (a ^ X & # 178;) & lt; loga (x + 1) & nbsp; = & gt; 0 & lt; a & lt; 1: A ^ X & # 178; & gt; X + 1; when a & gt; 1: A ^ X & # 178; & lt; X + 1
When 0 & lt; a & lt; 1, a ^ X & # 178; & gt; X + 1 & nbsp; = & gt; ln (a ^ X & # 178;) & gt; ln (x + 1) & nbsp; = & gt; & nbsp; X & # 178; LNA & gt; ln (x + 1), = & gt; LNA & gt; ln (x + 1) / X & # 178;;
Let f (x) = ln (x + 1) / X & # 178;, the derivative function is: [X-2 (x + 1) ln (x + 1)] / [(x + 1) x ^ 3], where x ∈ (0,1), the denominator is greater than 0, the derivative of the molecule is: - 1-2ln (x + 1) is less than 0, and the molecule is a decreasing function;, This is impossible
When a & gt; 1 has: A ^ X & # 178; & lt; X + 1, similar to 1. Has: LNA & lt; ln (x + 1) / X & # 178;, still let f (x) = ln (x + 1) / X & # 178;, then we should find the small value of F (x). F (x) is a decreasing function, when x = 1, we have the minimum value: LN2. But if x is less than 1, then f (x) = ln (x + 1) / X & # 178; & gt; LN2,
So LNA & lt; = LN2 & nbsp; = & gt; a & lt; = 2
So the value range of a is: & nbsp; 1 & lt; a & lt; = 2
It is known that two of the equations x Λ 2 + (M-17) x + (m-2) = 0 about X are positive real numbers. The value range of real number m is obtained
Because there are two roots of positive real numbers, this problem can be assumed as (x-a) * (X-B) = 0, where A.B is a positive real number, that is to say, X2 - (a + b) x + AB = 0 in this problem - (a + b) = M-17 AB = m-2 a + B = 17-m because AB is a positive real number, so 17-m & gt; 0 m & lt; 17 AB = m-2 & gt; 0 m & gt; 2, so this problem
Because there are two positive real roots, this problem can be assumed as (x-a) * (X-B) = 0, where A.B is a positive real number, that is to say, X2 - (a + b) x + AB = 0 in this problem - (a + b) = M-17 AB = m-2 a + B = 17-m because AB is a positive real number, so 17-m & gt; 0 m & lt; 17 AB = m-2 & gt; 0 m & gt; 2, so the solution to this problem is 2 & lt; M & lt; 17