In △ ABC, extend the midline BD on the AC side to F, so that DF = BD, extend the midline CE on the AB side to g, so that eg = CE

In △ ABC, extend the midline BD on the AC side to F, so that DF = BD, extend the midline CE on the AB side to g, so that eg = CE

Connecting CF and BG
AC and BF are equally divided
The quadrilateral abcf is a parallelogram
∴AF=BC
Similarly, Ag = BC
The proof of AF = AG
It is proved that if the limit of function f (x) exists when x → x0, then f (x) is bounded in a certain field at x0
The limit of function f (x) exists when x → x0. Let limf (x) = a (x → x0)
According to the definition: for any ε > 0, there exists δ > 0 such that when | x-x0|
Copy and paste a paragraph
Let x → x0, f (x) → a
Then for any ε > 0, there exists δ > 0
How to calculate the new coordinates when the point (x, y) rotates a degree around the origin
New coordinates (xcosa, ysina)
Linear Algebra: find the basic solution system of the following homogeneous linear equations: x1-2x2 + 4x3-7x4 = 0 2x1 + x2-2x3 + X4 = 0 3x1-x2 + 2x3-4x4 = 0
As shown in the figure, in the equilateral triangle ABC, D is the midpoint of AC, CE is the extension line of BC, and CE = CD, take the midpoint F of be, and verify that DF is vertical to be
prove:
Connect BD
∵ △ ABC is an equilateral triangle and D is the midpoint of AC
∴∠ACB=60°,∠BDC=30°
∵CD=CE
∴∠E=∠CDE
∵∠CDE+∠E=∠ACB=60°
∴∠E=30°
∴∠E=∠DBE
∴DB=DE
∵ f is the midpoint of be
Ψ DF ⊥ be (isosceles triangle with three lines in one)
It is proved that if the function is continuous in the interval [x0-a, x0], differentiable in (x0-a, x0), and limx - > x0 - (x0 left limit) f '(x) exists, then
Limx - > x0 - (left limit) f '(x) = left derivative at X0
This is the limit theorem of derivative, which can be proved by Lagrange formula
Let limx - > x0 - (x0 left limit) f '(x) = K
It is the left derivative of x0 at 00
So limx - > x0 - (left limit) f '(x) = x0 point left derivative
On the relationship between the new coordinate (x ', y') and the old coordinate (x, y) when the rectangular coordinate rotates around the origin
Let om = R. then x ′ = rcosc, y ′ = rsinc. X = rcosa = RCOs (c + b) = R [COSC · CoSb sinc · SINB] = x ′ cosb-y ′ sinby = rsina = rsin (c + b) = R [sinc · CoSb + COSC · SINB] = y ′ CoSb + X ′ SINB. The relationship between the new coordinates (x ', y') and the old coordinates (x, y) is: X
General solution of X1 + x2 + 2x3-x4 = 0 {- x-3x3 + 2x4 = 0 2x1 + x2 + 5x2-3x4 = 0
The format is {upper X1 + x2 + 2x3-x4 = 0 middle - x-3x3 + 2x4 = 0 lower 2x1 + x2 + 5x2-3x4 = 0
1 1 2 -1
-1 0 -3 2
2 1 5 -3
r2+r1,r3-2r1
1 1 2 -1
0 1 -1 1
0 -1 1 -1
r1-r2,r3+r2
1 0 3 -2
0 1 -1 1
0 0 0 0
The general solution of the equations is: C1 (- 3,1,1,0) ^ t + C2 (2, - 1,0,1) ^ t
In the triangle ABC, extend the midline BD and CE to f and G, so that DF = BD and eg = CE, and prove that G, a and F are collinear
It is proved that: connecting Ag and AF, because D is the midpoint of AC and E is the midpoint of AB, ED is the isosceles bisector of triangle CAG with GA as the bottom, so Ag / / ed, similarly, AF / / ed,
Because there can only be one line parallel to a line through a point, so g, a and F are collinear
Tongrentang trtgty6556535 dare not talk to him
In the proof of local boundedness of function limit, why is | f (x) - a | + | a | < a | + 1?
Do you mean by F (x) - a|
Using the inequality | x + y | ≤ | x | + | y|
LIM (x - > XO) f (x) = a for any ε > 0, there exists δ > 0, so that when | X - XO | < δ, there is always | f (x) - a | < ε
=>For ε 1 = 1, there exists δ 1 > 0, so that when | X - XO | < δ 1, there is always | f (x) - a | < ε 1
That is | f (x) - a | < 1
=>When | X - XO | < δ 1, | f (x) | < f (x) - a | + | a | < 1 + ||