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

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

, (vector DB + vector DC - vector 2da = (DB + DC + AD + AD) = (AC + AB)
(AB+AC)(AB-Ac)=0
ab=ac
an isosceles triangle
Y = x ^ 2-4 / x ^ 2-3x-2 the breakpoint type 'if it is removable' is redefined to be continuous
X = 0 is a removable breakpoint. Write another function y = 0, x = 0
Let {an} be a sequence of positive numbers, the sum of the first n terms is Sn, and for all n ∈ n +, the median of the equal difference between AM and 2 is equal to the median of the equal ratio between Sn and 2
(1) Write the first three terms of the sequence {an}
(2) Finding the general term formula of sequence {an} (writing out the reasoning process)
(2) 2,6,10 (2) from the meaning of the title, 2Sn = [(an + 2) / 2] square, Sn = an square / 8 + an / 2 + 1 / 2, then s (n-1) = a (n-1) square + a (n-1) / 2 + 1 / 2, the subtraction of the two formulas is: SN-S (n-1) = an = (an square-an-1 Square) / 8 + (an-an-1) / 2
It is known that A1, A2 and A3 are the three solutions of AX = B, and R (a) = 2, A1 = (1,1,1,), A2 + 3a3 = (3,2,1) find its general solution
Four different points a, B, C, D (DB + dc-2da) * (AB-AC) = 0 on the plane. The vector is in the bracket of the triangle ABC
Here is mainly vector operation
Because DB + dc-2da = (db-da) + (dC-dA) = AB + AC
So (DB + dc-2da) * (AB-AC) = 0, that is, (AB + AC) * (AB-AC) = 0
That is ab ^ 2-ac ^ 2 = 0, so AB = AC
So the triangle ABC is an isosceles triangle
The vector 2 words are omitted below
Take the midpoint P of BC, then Pb + PC = 0
DB+DC-2DA=(DB-DA)+(DC-DA)=AB+AC=(AP+PB)+(AP+PC)=2AP+(PB+PC)=2AP
AB-AC=CB
∴2AP*CB=0
That is AP ⊥ BC
｜ AB = AC (isosceles triangle with three lines in one)
The ABC is an isosceles triangle
According to the equation, DB + dc-2da = 0, because in the triangle ABC, DB + DC = BC, BC = 2da, with the triangle... can't...
The limit of 1 + X divided by n-th power and - 1 divided by n-th power, X tends to 0, and the limit of 1-x divided by - 3 divided by 2 + divided by 3-th power, X tends to - 5
Detailed problem solving process
The first problem: because when x approaches 0, according to the Equivalent Infinitesimal Substitution: (1 + x) ^ 1 / n-1 = x / n
So: LIM (x - > 0) [(1 + x) ^ 1 / n-1 divided by 1 / N] = 1
And the second question, because when x approaches - 5, the denominator of the numerator is not zero, so you can directly substitute - 5. If you feel ugly, you can further rationalize the denominator with the cubic difference formula, or you can directly put it there
The first one: one of the squares of n (using lobida's law)
The second one is to find the limit of elementary function.
The first problem uses the principle of infinitesimal equivalent substitution. Under the n-th root sign, 1 + X is equivalent to X / N, so the limit is 1
Just bring in the second question, ha ha
Let {an} be a sequence of positive numbers, the sum of the first n terms is Sn, and for all positive integers n, the median of the equal difference between an and 1 is equal to
The first three terms of {an} are
The ratio of the mean of an and Sn is equal to 1
(an +1)/2=√Sn
Sn=(an +1)²/4
When n = 1, S1 = A1 = (a1 + 1) & # / 4
(a1-1)²=0
a1=1
When n ≥ 2,
Sn=(an+1)²/4 Sn-1=[a(n-1)+1]²/4
Sn-Sn-1=an=(an+1)²/4 -[a(n-1)+1]²/4
(an-1)²=[a(n-1)+1]²
An - 1 = a (n-1) + 1 or an - 1 = - A (n-1) - 1 (an = - A (n-1), all items of the sequence are positive, rounding off)
an=a(n-1)+2
Sequence {an} is an arithmetic sequence with 1 as the first term and 2 as the tolerance
an=1+2(n-1)=2n-1
a1=1 a2=2×2-1=3 a3=2×3-1=5
Let A1, A2... As and B1, B2... BS be two linearly independent n-dimensional vector groups, and each A1 and B1 are orthogonal. It is proved that A1... As, B1... BS are independent
Let k1a1 +.. KSAs + M1b1 +.. + MSBS = 0, multiply M1b1 ^ t, m2b2 ^ t,. MSBS ^ t by left respectively, and then add them together
It is known that the first linear expression is independent of (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 + MBS) =... (M1 +
In the triangle ABC, given that the sum of vector AB and AC satisfies (the module of vector AB divided by ab plus the module of vector AC divided by AC), the point multiplication vector BC = 0, and the point multiplication vector AC divided by the module of vector AB divided by AC = two-thirds root sign, then what triangle ABC is?
I think the first condition can determine whether it is an isosceles triangle, plus the second condition or an isosceles triangle & nbsp;
Determine the second condition is the vector AB divided by the AB module point multiplication vector AC divided by the AC module = two-thirds root sign?
F (x) = x ^ k sin1 / X (x ≠ 0), 0 (x = 0) when k satisfies what condition, the function is (1) differentiable, (2) continuous, and (3) differentiable continuous when x = 0
Continuity must be equal to left and right neighborhoods, f (0) = 0f (0 +) = LIM (x - > 0 +) x ^ ksin1 / x = 0, k > = 1F (0 -) = LIM (x - > 0 -) x ^ ksin1 / x = 0, k > = 1. Therefore, when k > = 1, it is continuous at x = 0