If limf (x) = A and f (x) < 0, then A. > = 0, B. < = 0, C. > 0, D. < 0

If limf (x) = A and f (x) < 0, then A. > = 0, B. < = 0, C. > 0, D. < 0

Choose B
This is a corollary of local sign preservation

If f '(x0) = - 3, then limf (x0 + △ x) - f (x0-3 △ x) / △ x =?

The original formula = {[f (x0 + ⊿ x) - f (x0)] - [f (x0-3 ⊿ x) - f (x0)]} / ⊿ x
={[f(x0+⊿x)-f(x0)]/⊿x}+3{[f(x0-3⊿x)-f(x0)]/(-3⊿x)}
---->f'(x0)+3f'(x0)=4f'(x0)=-12

If f '(x0) = 2, then limf (x0 + 2 △ x) - f (x0-3 △ x) / △ x=

lim【△x→0】f(x0+2△x)-f(x0-3△x)/△x
=lim【△x→0】5[f(x0+2△x)-f(x0-3△x)]/(5△x)
=lim【△x→0】5[f(x0+2△x)-f(x0-3△x)]/[(x0+2△x)-(x0-3△x)]
=lim【△x→0】5f '(x0)
=5×2
=10
Answer: 10

The common tangent equation of two circles x ^ 2 + y ^ 2 = 1 and (x-3) + y ^ 2 = 4

One of the more intuitive is x = 3
Let the other two tangent equations be the intersection of y = KX + B and X axis
The centers of the two circles are B and C in turn
A (- 3,0) is obtained from similar triangles
The tangents y = (x + 3) / 2 and y = - (3 + x) / 2 are obtained in turn

As shown in the figure, it is known that triangle ABC is inscribed on circle O, point D is on the extension line of OC, ∠ ABC = ∠ CAD (1) judge the position relationship between straight line AD and circle O, and explain the reason
(2) If od is perpendicular to AB, BC = 5, ab = 8, find the radius of circle o

Given that ad is the height of △ ABC, ∠ bad = 62 ° and ∠ CAD = 28 °, what triangle is △ ABC?
Tip: there are two cases (right angle and obtuse angle)

Two cases
When ad is outside of △ ABC, it is an obtuse triangle
When ad is in △ ABC, ∠ BAC = 62 + 28 = 90 °, it is a right triangle

As shown in the figure, ⊙ o is the circumscribed circle of ⊙ ABC, FH is the tangent of ⊙ o, the tangent point is f, FH ∥ BC, the bisector BD connecting AF to e, the bisector BD connecting AF to D, connecting BF. (1) prove that AF bisects ∠ BAC; (2) prove that BF = FD; (3) if EF = 4, de = 3, find the length of AD

(1) FH is the tangent of \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\it's not easy In △ BFE and △ AFB, ∫ BDF = ∫ FBD, ∫ BF = FD (6 points) (3) in △ BFE and △ AFB, ∫ 5 = ∫ 2 = ∫ 1, ∫ AFB = ∫ AFB, ∫ BFE ∫ AFB (7 points) ∫ bfaf ∫ fefb, (8 points) ∫ BF2 = Fe · FA ∫ FA = bf2fe (9 points), EF = 4, BF = FD = EF + de = 4 + 3 = 7, ∫ FA = 724 = 494 ∫ ad = af-df = af - (de + EF) = 494 − 7 = 214 (10 points)

Circle O is the circumscribed circle of triangle, FH is the tangent of circle O, the tangent point is f, FH / / BC, the bisector connecting AF intersects BC at e, the bisector BD of angle ABC intersects AF at D and connects BF

It is proved that because FH is the tangent of a circle, the angle AFH = 90 degrees, because FH / / BC, the angle AEB = angle AFH = 90 degrees, so AE is perpendicular to BC, because AE passes through the center O, so be = CE, so BAE = angle CAE, so AF bisects the angle BAC

As shown in the figure, ⊙ o is the circumscribed circle of ⊙ ABC, FH is the tangent of ⊙ o, the tangent point is f, FH ∥ BC, the bisector BD connecting AF to e, the bisector BD connecting AF to D, connecting BF. (1) prove that AF bisects ∠ BAC; (2) prove that BF = FD; (3) if EF = 4, de = 3, find the length of AD

(1) It is proved that the connection of of ∵ FH is the tangent of ⊙ o ∵ of ⊥ FH (1 point) ∵ FH ∥ BC, ∵ of vertical bisection BC (2 points) ∵ BF = FC, ∵ 1 = ∵ 2, ∵ AF bisection ∵ BAC (3 points) (2) it is proved that from (1) and the conditions, we can know ∵ 1 = ∵ 2, ∵ 4 = ∵ 3, ∵ 5 = ∵ 2 (4 points) ∵ 1 + ∵ 4 = ∵ 2 + ∵ 3

Given the first term A1 = 3 of sequence {an}, and a (subscript n) - A (subscript n-1) = 2 (n is greater than or equal to 2), then a1 + A2 + a3=
I'm not sure which of the four numbers I calculated, 7, 15, 10, 8

This is an arithmetic sequence with 3 first term and 2 tolerance. The sum of the first three terms equals 15