If the corresponding points of two rational numbers on the number axis are on both sides of the origin, then the quotient of the two numbers A is a positive number, B is a negative number C is zero, D can be positive or negative

If the corresponding points of two rational numbers on the number axis are on both sides of the origin, then the quotient of the two numbers A is a positive number, B is a negative number C is zero, D can be positive or negative

B is negative!
It must be B
B
B. It's a negative number
If two of the equations x ^ 2-2ax + 4 = 0 are greater than one, then the value range of real number a is larger than one
Then the value range of real number a]
A more complete solution is as follows: let X &;, X &;, respectively. From the relationship between root and coefficient, it is obtained that: X &; + X &; = 2A, X &; * X &; = 4; the original equation has real roots, and both of them are greater than 1, then the following three conditions must be satisfied at the same time: ① △ = (- 2A) & sup2; - 4 × 4 ≥ 04A & sup2
In the inequality of X, loga (x2-4) > loga (6x-13a) (0
∵ loga (X & # 178; - 4) > loga (6x-13a), and a
It is known that the sum of the first n terms of the sequence an is Sn, and Sn = 10n-n & # 178;. Find the general term formula. Note BN = | an |, find the sum of the first n terms of the sequence {BN} TN
It is known that the sum of the first n terms of the sequence an is Sn, and Sn = 10n-n & # 178
Formula for finding general term·
Note BN = | an |, and find the sum TN of the first n terms of the sequence {BN}
When n = 1, A1 = S1 = 10-1 = 9
When n ≥ 2, an = SN-S (n-1) = 10n-n & # 178; - 10 (n-1) + (n-1) &# 178; = 11-2n
When n = 1, A1 = 11-2 = 9, which also satisfies the general formula
The general formula of sequence {an} is an = 11-2n
Let 11-2n ≥ 0
2n ≤ 11, n ≤ 11 / 2, and N is a positive integer, 1 ≤ n ≤ 5, that is, the first five items of the sequence are > 0, starting from the sixth item, and then all items are positive
From Sn = 10n-n2, sn-1 = 10 (n-1) - (n-1) 2, (n ≥ 2)
An = 11-2n can be obtained by subtracting the two formulas
When ∵ n = 1, A1 = S1 = 10-1 = 9
∴an=11-2n，∴bn=|11-2n|．
Obviously, when n ≤ 5, BN = an = 11-2n, TN = 10n-n2
When n ≥ 6, BN = - an = 2n-11,
Tn=（a1+a2+… +a5）-（a6+a7+… +Open up
From Sn = 10n-n2, sn-1 = 10 (n-1) - (n-1) 2, (n ≥ 2)
An = 11-2n can be obtained by subtracting the two formulas
When ∵ n = 1, A1 = S1 = 10-1 = 9
∴an=11-2n，∴bn=|11-2n|．
Obviously, when TN = 2n-10n = 5
When n ≥ 6, BN = - an = 2n-11,
Tn=（a1+a2+… +a5）-（a6+a7+… +an）=2S5-Sn=50-10n+n2
So TN=
10n-n2 (n ≤ 5) 50-10n + N2 (n ≥ 6) Stow
This is the third stage test paper (a) of the first semester of Grade 8 in Shanghai,
Given that a, B and C are the three sides of triangle ABC, please judge the equation 1 / 4x square - (a-b) x + C square = 0
Discriminant (a-b) ^ 2-4 * (1 / 4) * C ^ 2 = (a-b) ^ 2-C ^ 2
B + C > A and a + C > b
-c
For any real number a, the X equation x2-2ax-a + 2B = 0 has real roots, then the value range of real number B is ()
A. b≤0B. b≤−12C. b≤-1D. b≤−18
∵ the equation x2-2ax-a + 2B = 0 for X has real roots, ∵ a = 4a2-4 (- A + 2b) = 4a2 + 4a-8b = (2a + 1) 2-1-8b, for any real number a, ∵ a = (2a + 1) 2-1-8b ≥ 0, so - 1-8b ≥ 0, the solution is b ≤ − 18. So the value range of real number B is B ≤ − 18
Let f (x) = a − 22x + 1, where a is a constant; (1) f (x) is an odd function, try to determine the value of a; (2) if the inequality f (x) + a > 0 holds, find the value range of real number a
(1) ∵ f (x) is an odd function, ∵ f (- x) = - f (x), that is, A-22 − x + 1 = - A + 22x + 1, ∵ 2A = 22 − x + 1 + 22x + 1 = 2 · 2x1 + 2x + 22x + 1 = 2, ∵ a = 1; (2) f (x) + a ＞ 0 is constant, that is, a-22x + 1 + a ＞ 0, 2a ＞ 22x + 1 is constant, which is equivalent to 2A ＞ (22x + 1) max, while 2x ＞ 0, 2
The sum of the first n terms of the sequence {an} is Sn = 2n ^ 2 + 2n, and the sum of the first n terms of the sequence {BN} is TN = 2-BN
(1) Finding the general term formula of sequence {an} and {BN}
(1) When n = 1, S1 = 1-a1, so A1 = 1 / 2An = SN-S (n-1) = 1-an - (1-A (n-1) = a (n-1) - an, so: an = 1 / 2A (n-1), {an} is an equal ratio sequence, an = (1 / 2) ^ n (2) TN = 2 * 1 / 2 + 3 * (1 / 2) ^ 2 + +(n+1)*(1/2)^n1/2Tn=2*(1/2)^2+…… +N * (1 / 2) ^ n + (n + 1) * (1 / 2) ^ (n + 1) [this
S(n)=2*n^2+2n ①
S(n-1)=2*(n-1)^2+2(n-1) ②
① - 2
an=4n
T(n)=2-b(n) ①
T(n-1)=2-b(n-1) ②
① - 2
b(n）=b(n-1)-b(n)
B (n) =... Expansion
S(n)=2*n^2+2n ①
S(n-1)=2*(n-1)^2+2(n-1) ②
① - 2
an=4n
T(n)=2-b(n) ①
T(n-1)=2-b(n-1) ②
① - 2
b(n）=b(n-1)-b(n)
b(n)=b(n-1)/2 ③
It can be obtained from (1)
T(1)=2-b(1)
Namely
b(1)=2-b(1)
b(1)=1 ④
It can be obtained from (3) and (4)
B (n) = 2 ^ (- N + 1)
In the triangle ABC, the angle c is 90 degrees, the two right sides are a and B respectively, and a and B satisfy the equation A's Square - 3AB + 2B's Square = 0, so the value of sina can be obtained
How to divide the factorization factor of the square of a - 3AB + 2B = 0
From the equation a2-3ab + 2B = 0, (a-b) * (a-2b) = 0 can be calculated: a = B or a = 2B. When a = B, the triangle is isosceles right triangle, so Sina = sin45 degree is equal to two-thirds of the root sign. When a = 2B, C2 = A2 + B2, C2 = 4B2 + B2 = 5b2, so C is equal to root sign 5b, so Sina = 2B / root sign 5B = 5
B = 3 / 2A can be obtained by solving the equation
According to Pythagorean theorem, C = √ 13 / 2 a can be calculated
So Sina = A / C = √ 13 / 2
You're learning so fast. I haven't learned sin yet
On the inequality kx2-kx-1 of X
According to the meaning of the title:
When k = 0
The original inequality is - 1 < 0
When k ≠ 0
K < 0 (opening downward)
k²+4k＜0（△＜0）
Solution - 4 < K < 0
In conclusion - 4 < K ≤ 0
f(x) = kx2-kx-1
In order to make the parabolic opening upward when kx2-kx-10
f'(x) = 2kx -k = k(2x -1) = 0
x = 1/2
That is, when x = 1 / 2, f (x) = kx2-kx-1 is the extremum, f (1 / 2) = K / 4-K / 2-1 = - K / 4... Expansion
f(x) = kx2-kx-1
In order to make the parabolic opening upward when kx2-kx-10
f'(x) = 2kx -k = k(2x -1) = 0
x = 1/2
That is, when x = 1 / 2, f (x) = kx2-kx-1 is the extremum, f (1 / 2) = K / 4-K / 2-1 = - K / 4 - 1 < 0
k > -4
Combining with K < 0, we get - 4 < K < 0