The n-th power of a is B. how to find a with B and N The nth power of a = B. now we know B. how to find a?

The n-th power of a is B. how to find a with B and N The nth power of a = B. now we know B. how to find a?

(a^n)^(1/n)=b^(1/n)
So a = B ^ (1 / N)
On the inequality 2A ≤ x ≤ a ^ 2 + 1 of real number x and the solution set of x ^ 2-3 (a + 1) x + 2 (3a + 1) ≤ 0 (a ∈ R) are in turn a and B. find the value range of real number a that makes a subset of B
A={x|2a≤x≤a^2+1}
x^2-3(a+1)x+2(3a+1)≤0
(x-2)[x-(3a+1)]≤0
① When a < 1 / 3
B={x|3a+1≤x≤2}
Let a be a subset of B
So 3A + 1 ≤ 2a, 2 ≥ a ^ 2 + 1
So a = - 1
② When a ≥ 1 / 3
B={x|2≤x≤3a+1}
Let a be a subset of B
SO 2 ≤ 2a, 3A + 1 ≥ a ^ 2 + 1
So 1 ≤ a ≤ 3
So the value range of real number a is {a | 1 ≤ a ≤ 3 or a = - 1}
If you don't understand, please hi me, I wish you a happy study!
It is known that the equation MX & # 178; - MX + 2 = 0 about X has two equal real roots, so we can find the value of M
The solution of the equation MX & # 178; - MX + 2 = 0 has two equal real roots
Then m ≠ 0 and Δ = 0
That is, m ≠ 0 and Δ = (- M) &# 178; - 4 * 2 * m = 0
That is, m ≠ 0 and M & # 178; - 8m = 0
The solution is m = 8
What is the answer to the nth power of (a + b)?
The nth power of (a + b)
=C(n,0)a^(n)b^0+C(n,1)a^(n-1)b^1+C(n,2)a^(n-2)b^2+.+C(n,n)a^(0)b^n.
It is the binomial theorem
If the inequality 3x2 + 2x + 2x2 + X + 1 ＞ K holds for all real numbers x, the value range of K is obtained
If k = 3, the inequality is equivalent to - X-1 ＞ 0, that is, X ＜ - 1, which does not satisfy the condition. If K ≠ 3, to make the inequality constant, it is sufficient that 3 − K ＞ 0 △ = (2 − K) 2 − 4 (3 − K) (2 − K) ＜ 0, that is, K ＜ 3 (K − 2) (3K − 10) ＞ 0, ‖ K ＜ 3K 103 or K < 2, i.e. K < 2
If the equation MX & # 178; + (2m + 1) x + M = 0 has only one real root, then the value of the real number m is zero
If one of the quadratic equations ax & # 178; - 5x + A & # 178; + a = 0 is followed by 0, then the value of a is
If the equation MX & # 178; + (2m + 1) x + M = 0 has only one real root
Then: △ = (2m + 1) &# 178; - 4m & # 178; = 0
4m+1=0
m=-1/4
In addition, when m = 0, the equation is x = 0, and there is only one real root
So: M = - 1 / 4 or M = 0
If one of the quadratic equations ax & # 178; - 5x + A & # 178; + a = 0 with respect to x follows 0
Substituting x = 0 gives:
A & # 178; + a = 0; and a ≠ 0
a(a+1)=0
a=-1
Because the equation MX & # 178; + (2m + 1) x + M = 0 has only one real root
1) When m is not equal to 0, then
So the discriminant of root B ^ 2-4ac = 0
(2m+1)^2-4m^m=0
4m+1=0
m=-1/4
2) When m = 0, substitute it into the original equation, 0 = 0, the equation holds
So m = 0 holds
Because, one of the univariate quadratic equations ax & # 178; - 5x + A & # 178; + a = 0 of X is expanded with
Because the equation MX & # 178; + (2m + 1) x + M = 0 has only one real root
1) When m is not equal to 0, then
So the discriminant of root B ^ 2-4ac = 0
(2m+1)^2-4m^m=0
4m+1=0
m=-1/4
2) When m = 0, substitute it into the original equation, 0 = 0, the equation holds
So m = 0 holds
Because one of the quadratic equations ax & # 178; - 5x + A & # 178; + a = 0 of X follows 0,
So, a is not equal to 0. Substituting x = 0 into the original equation, we get
a^2+a=0
a+1=0
A = - 1 ﹣ put away
The first case
When the equation is univariate: M = 0
The equation has only one real root
The second situation
When the equation is quadratic
Only △ = 0
△=（2m+1)(2m+1)-4m*m
=4m+1
=0
Then M = - 1 / 4
{2x + 5Y = 8,3x + 2Y = 5 by substitution
The results are as follows
x=4-2.5y ③
Substituting in (2)
3（4-2.5y）+2y=5
12-5.5y=5
7=5.5y
The solution is y = 14 / 11
Substituting 3, we can get the following solution
X = 9 / 11
The value range of real number a when inequality system | x-a | = 0
X ^ 2-5x + 4 > = 0
X ≤ 1, or X ≥ 4
And the system of equations has no solution
1+1
For x ^ 2-5x + 4 ≥ 0, (x-1) (x-4) ≥ 0
Either: X-1 ≥ 0, x-4 ≥ 0, the solution is: X ≥ 4, that is: X ∈ [4, ∞);
Or: X-1 ≤ 0, x-4 ≤ 0, the solution is: X ≤ 1, that is: X ∈ (- ∞, 1]
Obviously, if we want the inequality to have a solution, X cannot be in the interval (1,4).
Look again: | x-a | ≤ 1
When x ≥ a, | x-a | ≤ 1, sorted as: x-a ≤ 1, the solution is: X ≤ a + 1, X ∈ (- ∞, a + 1]; = 0
have to
X ≤ 1, or X ≥ 4
We can see from the picture
When there is no solution
A-1 > 1 and a + 1
If the equation MX & # 178; - MX + 2 = 0 has two equal real roots, then M______ .
I will hand it in at 2:00 this afternoon. I hope I can get your help as soon as possible!
If there are two equal real roots, then △ = 0
m²-8m=0
m(m-8)=0
m=0,m=8
There are two equal real roots,
So it's a quadratic equation of one variable
So the coefficient of X & # is not equal to 0
therefore
M=8
m≠0
m^2-8m=0
m(m-8)=0
M = 0 or M = 8
So, M = 8
Eight
When x is equal to 3, 2x + 9y = 56 3x + 5Y + 2 + 68 ask: what is x and y equal to
"*" is the multiplication sign
2*3=6 6+9y=56 9y=56-6 9y=50
3 * 3 = 9 9 + 5Y + 2? 68 "?" is "="?
Is the title wrong?
I don't know. Ha ha
Is it lack of known condition that we can only get a ratio of X and y