On the number axis, point a (representing integer a) is on the left side of origin o, and point B (representing integer b) is on the right side of origin O. if A-B = 2013 and AO = 2BO, then the value of a + B is () A. -1242B. 1242C. 671D. -671

On the number axis, point a (representing integer a) is on the left side of origin o, and point B (representing integer b) is on the right side of origin O. if A-B = 2013 and AO = 2BO, then the value of a + B is () A. -1242B. 1242C. 671D. -671

As shown in the figure, a ＜ 0 ＜ B. ∵ | A-B | = 2013, Ao = 2BO, ∵ B-A = 2013, ① a = - 2b, ② from ① and ②, B = 671, ∵ a + B = - 2b + B = - B = - 671
As shown in the figure, eight points a, B, C, D, e, F, G, h on the number axis represent integers. If the number corresponding to B is B, the number corresponding to e is e, and e-2b = 7, then the origin of the number axis is______ Point
According to the position of the point on the number axis, we can express the relationship between E and B: e = B + 3, and e-2b = 7, then B + 3-2b = 7, ∧ B = - 4. That is, B is on the left side of the number axis, four unit lengths away, and the origin is f
Compare the size of 1111 out of 1111 and 1111 out of 11111
subject
From 11111 / 1111 = 10 and 1 / 11111 / 111 = 10 and 1 / 111 10 and 1 / 1111
1+11+111+1111+11111+…… +11…… 11 (2009 1) and how many 1?
223 1
1 + 2 + 3 +... + 2009 = (1 + 2009) / (2009 / 2) = 2019045?
If we know the quadratic function f (x) = x2 + ax + B, if the inequality f (x) about X
The solution is x2 + ax + B
The general term formula of an is an = 2N-1 sequence BN = 1 / (Anan + 1), the first n terms and Sn are equal
bn=1/(2n-1)(2n+1)
=1/2*2/(2n-1)(2n+1)
=1/2*[(2n+1)-(2n-1)]/(2n-1)(2n+1)
=1/2*[(2n+1)/(2n-1)(2n+1)-(2n-1)/(2n-1)(2n+1)]
=1/2[1/(2n-1)-1/(2n+1)]
So Sn = 1 / 2 * [1 / 1-1 / 3 + 1 / 3-1 / 5 + +1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)
Verification: (3-sin ^ 4 α - cos ^ 4 α) / 2cos ^ 2 α + 1 + Tan ^ 2 α + sin ^ 2 α
Sorry, it should be
Prove: (3-sin ^ 4 α - cos ^ 4 α) / (2cos ^ 2 α) = 1 + Tan ^ 2 α + sin ^ 2 α
Proof: let x = (COS α) ^ 2,
Then (sin α) ^ 2 = 1-x,
(tan α)^2 =(1-x) /x.
So left = [3 - (1-x) ^ 2 - x ^ 2] / (2x)
=(2 +2x -2x^2) /(2x)
=(1 +x -x^2) /x,
Right = 1 + (1-x) / x + (1-x)
=(1 +x -x^2) /x.
So left = right,
The original equation holds
= = = = = = = = =
Exchange method
Pay attention to the number of times
(sin α) ^ 4 = (1-x) ^ 2, not (1-x) ^ 4
In fact, trigonometric function problem
3=2+1=2+(sin^2α+cos^2α)^2=2+2sin^2αcos^2α+sin^4α+cos^4α
Therefore, 3-sin ^ 4 α - cos ^ 4 α = 2 + 2Sin ^ 2 α cos ^ 2 α
So: left = (2 + 2Sin ^ 2 α cos ^ 2 α) / (2cos ^ 2 α) = 1 / cos ^ 2 α + sin ^ 2 α = (sin ^ 2 α + cos ^ 2 α) / cos ^ 2 α + sin ^ 2 α
=1+tan^2α+sin^2α
It is proved.
Resolve or verify? You don't have an equation to prove it. Follow up: I just dialed the wrong number. Please help me. Thank you
Discuss the function f (x) = ax / x2-1 (- 1)
X1
The approximate interval of the zero point of the function f (x) = LNX − 2x is ()
A. (1，1e)B. （e，+∞）C. （1，2）D. （2，3）
For the function f (x) = LNX − 2x is continuous on (0, + ∞), because f (2) = ln2-22 ＜ 0, f (3) = ln3-23 ＞ 0, so f (2) f (3) ＜ 0, so the approximate interval of zero point of function f (x) = LNX − 2x is (2,3), so D
If the sum of the first n terms of the arithmetic sequence {an}, {BN} is Sn, TN respectively, and sntn = 3N − 12n + 3, then a8b8=______ .
2a82b8 = a1 + a15b1 + B15 = 152 (a1 + A15) 152 (B1 + B15) = s15t15 = 3 × 15 − 12 × 15 + 3 = 43