150 mathematical problems Please make a question

150 mathematical problems Please make a question

It is suggested to refer to books and matching exercise books
Compare the following numbers: 2, radical 5, 3 ^ radical 7
√5≈2.236…
3^2
Grade one mathematics 2 * square root 3 * square root 5
Thank you. It's better to have a process. Well, thank you
2×√3×√5=2×√(3×5)=2√15
Original formula = 2 √ 15
2×√3×√5=2×√(3×5)=2×√15=7.745966692414
I'm glad to answer for you. I wish you progress in your study!
If the domain of function f (x) = LG (KX ^ 2-2kx + 2 + k) is r, then the range of real number k is
(1)
If k = 0
The true number n = 2 > 0 satisfies the meaning of the question;
(2)
If K ≠ 0, then
{k>0
{Δ=4k^2-4k(k+2)= - 8k0
In conclusion, K ≥ 0
K (x ^ 2-2x + 1) + 2 > 0 to simplify K (x-1) ^ 2 + 2 > 0 x belongs to R. the value range of K is (negative infinity to zero)
Domain: the value range of the independent variable that makes the function meaningful.
LG (T) is significant when t > 0
The domain KX ^ 2-2kx + 2 + k > 0 of the topic function is equivalent to x belonging to R
That is to say, constant is greater than 0
Then, when k = 0, it holds
When k is not equal to 0, and the quadratic function is always greater than 0, then k > 0, and the minimum x = 1, K.... > 0, the inequality k > 0 is solved
To sum up, k > = 0
Domain: the value range of the independent variable that makes the function meaningful.
LG (T) is significant when t > 0
The domain KX ^ 2-2kx + 2 + k > 0 of the topic function is equivalent to x belonging to R
That is to say, constant is greater than 0
Then, when k = 0, it holds
When k is not equal to 0, and the quadratic function is always greater than 0, then k > 0, and the minimum x = 1, K.... > 0, the inequality k > 0 is solved
To sum up, k > = 0
It holds when k = 0.
When k ≠ 0, if KX ^ 2-2kx + 2 + k > 0 is always true, the image opening is upward, and △ 0 (2k) ^ 2-4k (2 + k) 0
So K ≥ 0
If y = x ^ 2-2ax + 3 (1
If y = x ^ 2-2ax + 3 (1
The function f (x) = y = LG (x ^ 2-kx-k) is known. 1. The domain of definition is r, and the range of value of K is obtained. 2. The domain of value is r, and the range of value of K is obtained
1. The domain of definition is r
Discriminant = k ^ 2 + 4K
（-4,0）
1, the definition field is r, meaning x can take any real number, f (x) has a value, so the requirement: x ^ 2-kx-k > 0, that is, the discriminant 0, is the same as the answer to question 1, but the question does not require that it must be meaningful, so the final answer is (- infinite, +... Expansion
1, the definition field is r, meaning x can take any real number, f (x) has a value, so the requirement: x ^ 2-kx-k > 0, that is, the discriminant 0, is the same as the answer to question 1, but the question does not require that it must be meaningful, so the final answer is (- infinity, + infinity)
What is the inverse of the function f (x) = LNX
y=lnx
e^y=x
x=e^y
So the inverse of the original function is:
y=e^x
The x power of e
If the domain of F [LG (x + 1)] is [0.9], then the domain of F (x of 2) is?
X belongs to [0,9]
X + 1 belongs to [1,10]
LG (x + 1) belongs to [0,1]
So f (), this bracket can only be [0,1]
That is to say, 2 ^ x belongs to [0,1]
Then x belongs to (negative infinity, 0]
The domain is (negative, infinite, 0]
Given that f (x) = x LNX, when a > 0, b > 0, we prove that f (a) + F (b) > = f (a + b) - (a + b) LN2
f(a+b)-(a+b)ln2=2f[(a+b)/2]
f(x)'=lnx+1
f(x)''=1/x
X > 0
∴f(x)''>0
F (x) is a lower convex function
According to the mean value theorem, the lower convex function [f (a) + F (b)] / 2 > = f [(a + b) / 2]
∴f(a)+f(b)>=f(a+b)-(a+b)ln2
Get proof
prove:
When x > 0, f '' (x) = 1 / x > 0,
F (x) is a concave function on (0, + ∞)
1) If a ≠ B, let AF [(a + b) / 2]
Namely
[alna + blnb] / 2 > [(a + b) / 2] × ln [(a + b) / 2]
Alna + blnb > (a + b) ln [(a + b) / 2] = (a + b) ln (a + b) - (... Expansion)
prove:
When x > 0, f '' (x) = 1 / x > 0,
F (x) is a concave function on (0, + ∞)
1) If a ≠ B, let AF [(a + b) / 2]
Namely
[alna + blnb] / 2 > [(a + b) / 2] × ln [(a + b) / 2]
alna+blnb>(a+b)ln[(a+b)/2]=(a+b)ln(a+b)-(a+b)ln2
That is f (a) + F (b) > F (a + b) - (a + b) LN2
2) When a = B,
f(a)+f(b)=2f(a)=2alna
f(a+b)-(a+b)ln2=f(2a)-2aln2=2aln(2a)-2aln2=2aln(2a/2)=2alna
So f (a) + F (b) = f (a + b) - (a + b) LN2
Therefore, from 1) 2) it can be concluded that:
f(a)+f(b)>=f(a+b)-(a+b)ln2
Put it away
Given the domain of function y = f (x), what is the domain of function g (x) = f (x ^ 2) / LG (x-1) + 1 in (0,3)
Why the last x
Y = f (x) X domain (0,3)
Then f (x ^ 2) x ^ 2 ∈ (0,3) so - root 30 so x > 1
The denominator is not equal to 0, and the denominator is LG (x-1), so x ≠ 2
So the domain is x ∈ (1, root 3)