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

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

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Judge the type of F (x) = (1 + x) / SiNx discontinuity
X = 0 is the removable discontinuity of the first type, and x = k is the second type (k)
Sum of the first n terms in the equal ratio sequence Sn sum of the reciprocal of the first n terms TN for Sn / TN
Let the first term of the equal ratio sequence be a and the ratio be Q
Then Sn = a (1-Q ^ n) / (1-Q)
The first term of the countdown sequence is 1 / A, and the ratio is 1 / Q
Tn={1/a[(1-(1/q)^n)]}/(1-1/q)=q(q^n-1)/[aq^n(q-1)]
Sn/Tn=a^2q^(n-1)
That is, the product of the first term and the last term of the original equal ratio sequence
Equivalent to an
Let A1 = (1,2,3,3), A2 = (1,3,3,4), A3 = (1,2,3,4) be the three solution vectors of the four variable non-homogeneous linear equations with coefficient matrix A
And R (a) = 2
Because the rank of a is 2, ax = 0 has two linearly independent solutions: B1 = a1-a2 = (0, - 1,0, - 1), B2 = a1-a2 = (0,0,0, - 1)
The solution of this non second order system is x = K1 * B1 + K2 * B2 + a3, where K1 and K2 are arbitrary constants
As shown in the figure, D is a point within △ ABC, and the extension line of DB = DC, ab = AC, ad intersects BC at point E. verify AE ⊥ BC
It is proved that: DB = DC, ab = AC means that D A is on the vertical bisector of line BC, so ad is perpendicular to BC, the extension of ad intersects BC at e, and E is on ad, so a e ⊥ BC
Find the discontinuous point of F (x) = (x + 1) SiNx / | x | (x + 1) (x-1), and distinguish the type
The discontinuity is x = 0, x = 1, x = - 1
f(-x)=(-x+1)sin(-x)/|x|(-x+1)(-x-1)=(x-1)sinx/|x|(x-1)(x+1)=\=f(x) or -f(x)
So it's not odd or even
Let TN, Sn be the reciprocal of TN, Sn}
Sn = A1 (1-Q ^ n) / (1-Q) TN = 1 / A1 (1-1 / Q ^ n) / (1-1 / Q) PN = A1 ^ n * q ^ (n * (n-1) / 2) the square of PN under N times the square of PN under TN = SNN times the square of PN under N times the root = A1 ^ 2 * q ^ (n-1) A1 ^ 2 * q ^ (n-1) * TN = A1 (Q ^ (n-1) * - 1 / Q) / (1-1 / Q) = A1 (1-Q ^ n) / (1-Q) = Sn when Q = 1, Sn = na1tn = n / a1pn
There is a column of numbers A1, A2, A3,... An Starting from the second number, each number is equal to the difference between 1 and the reciprocal of the number before it. If A1 = 2, what is the value of A2009?
A: (- 1) B: 1 C: 3 D: 3 or - 1
A: (- 1) B: 1 C: 3 D: 3 or - 1
There is no such thing!
It should be a: 2009 B: 2 C: 1 / 2 D: - 1
2 1/2 -1 2 1/2 -1 2 1/2 -1 2 .
2009 △ 3 = 669 + 2
So the answer is 1 / 2
In order to know that A.B.C is the three sides of △ ABC, and satisfy a ^ 2 + 2B ^ 2 + C ^ 2-2b (a + b) = 0, try to judge the shape of the triangle
a^2+2b^2+c^2-2b(a+c)
=(a^2+b^2-2ab)+(b^2+c^2-2bc)
=(a-b)^2+(b-c)^2=0
So: a = B, B = C
That is, a = b = C
This triangle is equilateral
F (x) = ((e ^ SiNx) - 1)) / X find the break point and judge the type