![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Ad is the height of the side BC of the triangle ABC. Take ad as the diameter to make a circle, intersect with AB and AC respectively, multiply AC by E and F, and prove that AE multiplied by ab equals AF
Proof: link ed, FD
Because ad is the diameter
Therefore, AED = AFD = 90 degree
Because ad ⊥ BC
So △ AED ∽ ADB, △ AFD ∽ ADC
So AE: ad = ad: AB, AF: ad = ad: AC
So AE * AB = ad * ad, af * AC = ad * ad
So AE * AB = af * AC
Connect ed, FD
Because ad is the diameter
Therefore, AFD = 90 ∠
Because ad BC
So AE * AB = ad * ad, af * AC = ad * ad
So AE * AB = af * AC
The following functions are discontinuous at the points indicated, indicating which class these discontinuous points belong to. If they are removable discontinuous points, the definition of functions shall be supplemented or explained to make them continuous
(1)y=x^2-1/x^2-3x+2,x=1,x=2
(2)y=x/tanx,x=kπ,x=kπ+π/2(k=0,±1,±2...)
(3)y=cos^2 1/x,x=0
1、y=(x-1)(x+1)/[(x-1)(x-2)],
When x = 1, Lim [x → 1] (x-1) (x + 1) / [(x-1) (X-2)] = Lim [x → 1] (x + 1) / (X-2) = - 2,
If x = 2 ∫, Lim [x → 2] (x-1) (x + 1) / [(x-1) (X-2)] = ∞,
X = 2 is an infinite discontinuous point, belonging to the second kind of discontinuity,
If f (1) = - 2, then x = 1 is continuous, so x = 1 is a removable discontinuity
2. When x = k π (K ≠ 0), the denominator is 0, which is the second kind of discontinuity,
But if k = 0, Lim {x → 0) (x / TaNx) = 1, the limit exists. If f (0) = 1, it is a continuous point, so it belongs to a removable discontinuity,
When x = k π + π / 2, Lim {x → K π + π / 2) (x / TaNx) = 0, can complement f (K π + π / 2) = 0, so it belongs to removable discontinuity
3、y=cos^2（ 1/x）[1+cos(2/x)]/2,
X = 0, the denominator is 0, which is the breakpoint. Lim {x → 0) [cos ^ 2 (1 / x)] does not exist, which belongs to the second type of breakpoint
What do you mean? Follow up: ·························=
Let the sum of the first n terms of the sequence {an} be Sn and satisfy S1 = 2, Sn + 1 = 3Sn + 2 (n = 1, 2, 3,...) (I) prove that the sequence {an} is an equal ratio sequence and find the general term an; (II) find the first n terms and TN of the sequence {Nan}
It is proved that: (1) by subtracting ∵ Sn + 1 = 3Sn + 2, ∵ Sn = 3sn-1 + 2 (n ≥ 2), an + 1 = 3an (n ≥ 2) ∵ S1 = 2, Sn + 1 = 3Sn + 2 ∵ a1 + A2 = 3A1 + 2, that is, A2 = 6, then a2a1 = 3 ∵ an + 1An = 3 (n ≥ 1) ∵ sequence {an} is an equal ratio sequence with the first term of 2 and the common ratio of 3 ∵ an = 2 × 3n-1 (n = 1, 2, 3,...) ．（Ⅱ）∵Tn=1•a1+2•a2+… +nan=1×2+2×2×31+… +n×2×3n-1，∴3Tn=1×2×3+2×2×32+… +(n-1) × 2 × 3n-1 + n × 2 × 3N, (9 points); - 2tn = 2 (1 + 3 + 32 +...) 3n-1) - n × 2 × 3N = 2 × 3N − 13 − 1-N × 2 × 3N = 3N (1-2n) - 1 (11 points) ‖ TN = (2n − 1) 3N + 12 & nbsp; (13 points)
In the arithmetic sequence {a}, if A3 + A4 + A5 = 12, a1 + A2 +... + A7 is equal to
Because {an} is an arithmetic sequence
So there is: a1 + A7 = A3 + A5
a2+a6=a3+a5
2a4=a3+a5
Because A3 + A4 + A5 = 12
So 3A4 = 12
a4=4
a3+a5=8
So a1 + A7 = A2 + A6 = 8
So a1 + A2 + a3 + A4 + A5 + A6 + A7 = 8 + 8 + 8 + 4 = 28
Let the tolerance of arithmetic sequence an be D, and the result is as follows
a3=a4-d；a5=a4+d
a3+a4+a5=(a4-d)+a4+(a4+d)=3a4=12
Similarly, A1 = a4-3d, A2 = a4-2d ，a7=a4+3d；
a1+a2+a3+a4+a5+a6+a7=7a4=28
a3+a4+a5=3a4=12 a4=4 a1+a2...+a7=7a4=28
It is known that ad is the middle line on the side BC of △ ABC, G is the center of gravity of the triangle, EF passes through point G and is parallel to BC, and intersects AB and AC at points E and f respectively. The values of AF: FC and EF: BC are obtained
∵ G is the center of gravity of the triangle, and ad is the middle line on the edge of BC, ∵ Ag: GD = 2:1, Ag: ad = 2:3, ∵ EF ‖ BC, ∵ AF: FC = Ag: GD = 2:1, EF: BC = AF: AC = Ag: ad = 2:3
The function is discontinuous at a given point, which class these discontinuous points belong to
If it is a removable breakpoint, the definition of the function is supplemented or changed to make the function continuous at that point
Y = x ^ 2-1 divided by x ^ 2-2x-3 x = - 1 x = 3
The teacher didn't understand this question in class. Please tell me the way of thinking after solving this question
If there exists and the left and right limits are equal, then it is a removable discontinuity. A variant of the formula is y = (x + 1) (x-1) / (x + 1) (x-3). When x approaches to 3, the molecule approaches to constant 8, and
Let the sum of the first n terms of the sequence {an} be Sn, if S1 = 1, S2 = 2 and S (n + 1) - 3Sn + 2S (n-1) = 0 (n ≥ 2 and N ∈ n +), try to judge {an}
Is it an equal ratio sequence?
a1=s1=1 a2=s2-s1=1
S (n + 1) - 3Sn + 2S (n-1) = 0 (n ≥ 2 and N ∈ n +),
S (n + 1) - Sn = 2 [SN-S (n-1)], that is, a (n + 1) = 2An
And A2 / A1 = 1 is not equal to 2
So {an} is not an equal ratio sequence
(except for the first term, it will be an equal ratio sequence)
Solve the equations (x1 + 2x2 + 2x3 + X4 = 0, 2x1 + x2-2x3-2x4 = 0, x1-x2-4x3-3x4 = 0)
x1+2x2+2x3+x4=0 (1) 2x1+x2-2x3-2x4=0 (2) x1-x2-4x3-3x4=0 (3)(2)-(3)x1+2x2+2x3+x4=0 = equation (1)rank of system of equations = 2(1)+(2)3x1+3x2+3x4=0x4=-(x1+x2)from (1)x1+2x2+2x3-(x1+x2)=0x3 = -x2/2sol...
X1=1，X2=-1，X3=0.5，X4=0
It can be brought in for checking
Indefinite solution
The solutions of the equations are: (16,9, - 6,0) &# - 39; + C (15,8, - 5,1) &#. X1-3x2-2x3-x4 = 1 (1) 3x1-8x2-4x3-x4 = 0 (2) - 2x1 + x2-4x3 + 2x4 = 1
Point O is any point on the middle line ad of triangle ABC. The extension lines of Bo and co intersect AC and ab at points E and f respectively, and connect EF
Auxiliary line: through o as parallel line of AB, intersection AB to g, intersection AC to h △ EBC ∽ EOH, be / OE = Oh / BC ∽ DCB ∽ dog, CD / OD = og / BC, because Oh = og, be / OE = CD / OD, that is, (OE + OB) / OE = (OD + OC) / OD, ob / OE = OC / OD, because ∠ DOE = ∠ BOC, so △ doe ∽ cob, so ∠ ode = ∠ OCB, so
Why don't you give me a picture
= =
The proof of the first kind of functions with original points
Suppose there is a primitive function f (x), the primitive function is continuous, and C is the first kind of discontinuity of F (x), then f (c) is the derivative of the primitive function at x = C. at the same time, f (x) should be continuous in the field of C. this is contrary to the assumption that x = C is the first discontinuity of F (x). Therefore, there is no primitive function
In the first kind of discontinuity, X exists, but only two corresponding y are jumping discontinuities or this point is meaningless. However, if there is an original function, y must exist and be unique
Continuous function has primitive function, the function of the first kind of discontinuous point is not continuous function, so it has no primitive function. How to prove that it is not a continuous function?
RELATED INFORMATIONS
- 1. In the known sequence an, A1 = 2, the sum of the first n terms is SN. For any n ≥ 2, 3sn-4, an, 2-3 / 2S (n-1) is an arithmetic sequence (1) (2) if BN satisfies BN = 3Sn, find the first n terms and TN of BN
- 2. If AQ = 1 / mAb, AP = 1 / NAC, then M + n is equal to A.1 B.2 C.3 D.4
- 3. It is known that the equal ratio sequence satisfies A1 = 2, A2 + a3 = 12 and A4 > 0.1. Find the general term formula of sequence an. 2. If BN = anlog2an, the first n term of sequence BN is SN The main thing is to ask the second question!
- 4. In the RT triangle ABC, the angle c = 90, BC = a, AC = B, a + B = 16, then the function analysis of the area s of the RT triangle with respect to the variable length a When s = 32, a=______
- 5. The N-1 power {BN} of an and Sn = 2-1 / 2 is the arithmetic sequence A1 = B1, A2 * (b2-b1) = A1 to find the general term of BN? Let CN = BN / an to find the general term of CN
- 6. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ AB is at e, DF ⊥ AC is at F, and be = CF
- 7. In the arithmetic sequence {an}, if A1 = 3, A100 = 36, then A37 =?
- 8. As shown in the figure, BD and CE are the two heights of triangle ABC, and m and N are the midpoint of BC and de respectively
- 9. It is known that [an] and {BN} are arithmetic sequences, and CN = 3an + 2. It is proved that {CN} is also arithmetic sequences (2) The second question CN = an + BN (AB is not equal to 0): the sequence {CN} is an arithmetic sequence
- 10. In the RT triangle ABC, the angle BAC = 90, ad ⊥ BC at point D, e is the midpoint of AD, the extension line connecting ed and intersecting AB at point F, AB / AC = DF / AF is proved E is the midpoint of AC
- 11. Let BN = an + n prove that {BN} is an equal ratio sequence
- 12. In the triangle ABC, D is the midpoint of BC, AE bisector angle BAC, make de perpendicular to AE at e, intersect AB at g, intersect AC extension line at h. proof: BG = ch = 1 / 2 (AB-AC)
- 13. In the equal ratio sequence {an}, for any natural number n, there are a1 + A2 + +An = 2 ^ n-1, then (A1) ^ 2 + (A2) ^ 2 + +（an）^2=?
- 14. A. If B, C and D are four non collinear points, (vector DB + vector DC vector 2da) * (vector AB vector AC) = 0, then the shape of triangle ABC is
- 15. Let {an} be a sequence of positive terms, the sum of the first n terms is Sn, and for all positive integers n, the mean of the equal difference between an and 2 is equal to the mean of the equal ratio between Sn and 2 (1) Find the first three terms of the sequence; (2) find the general term formula of the sequence {an}
- 16. In the known triangle ABC, a (2, - 1), B (3,2), C (- 3, - 1), the height of the edge of BC is ad
- 17. In the equal ratio sequence {an}, the first term A1 ＜ 0, if the sequence {an} has an + 1 ＞ an for any positive integer n, then the common ratio Q should satisfy A.q＞1 B.0＜q＜1 C.1/2＜q＜1 D.-1＜q＜0
- 18. If the area of △ ABC is known as s, and the vector ab ● vector BC = 1, if s = 3 / 4 | vector ab |, find the minimum value of | vector AC |
- 19. In the equal ratio sequence {an}, A1 = 512, common ratio q = negative half, and TN is used to express the product of the first n terms; In the equal ratio sequence {an}, A1 = 512, common ratio q = negative half, and TN is used to represent the product of his first n terms; TN = A1A2... An, then the largest of T1, T2,... TN is:
- 20. As shown in the figure, △ ABC, ab = AC, ad ⊥ BC intersection D point, e and F are the midpoint of DB and DC respectively, then the logarithm of congruent triangle in the figure is () A. 1B. 2C. 3D. 4