I'll let you draw the figure.

a) AM is a median of the triangle ABC. Therefore, M is the midpoint of BC.

E is symmetric to A through M, so M is the midpoint of AE.

Consider the quadrilateral ABEC, it has M is the midpoint of both AE and BC. Thus, ABEC is a parallelogram.

b) Consider the triangle ADE, M is the midpoint of AE, H is the midpoint of AD. Therefore, HM is the average line of this triangle. This means \(HM//DE\) or \(DE//BC\), which means BCED is a trapezoid.

Triangle ABD has the height BH, which is also a median. Thus, ABD must be an isosceles triangle, which means the height BH is also a bisector, or \(\widehat{ABH}=\widehat{DBH}\) or \(\widehat{ABC}=\widehat{DBC}\)

On the other hand, ABEC is a parallelogram. So, \(AB//CE\) and this leads to \(\widehat{ABC}=\widehat{ECB}\) (2 staggered angles of 2 parallel lines)

From these, we have \(\widehat{DBC}=\widehat{ECB}\left(=\widehat{ABC}\right)\)

The trapezoid BCED (DE//BC) has \(\widehat{DBC}=\widehat{ECB}\). Therefore, BCED must be an isosceles trapezoid.

tính nhanh : 98.2 + 96.4 + 961.8

= 98.2 + 961.8 + 96.4

=(98.2 + 961.8) + 96.4

=1060+96.4

=1256.4

\(B=\dfrac{-1}{4}+\dfrac{7}{33}-\dfrac{5}{3}-(\dfrac{-15}{12}+\dfrac{6}{11}-\dfrac{68}{49})\)

\(B=\dfrac{-1}{4}+\dfrac{7}{33}-\dfrac{5}{3}+\dfrac{15}{12}-\dfrac{6}{11}+\dfrac{68}{49}\)

\(B=(\dfrac{-1}{4}-\dfrac{5}{3}+\dfrac{15}{12})+(\dfrac{7}{33}-\dfrac{6}{11})+\dfrac{68}{49}\)

\(B=\dfrac{-3-20+15}{12}+\dfrac{7-18}{33}+\dfrac{68}{49}\)

\(B=\dfrac{-8}{12}+\dfrac{-11}{33}+\dfrac{68}{49}\)

\(B=\dfrac{-2}{3}+\dfrac{-1}{3}+\dfrac{68}{49}\)

\(B=-1+\dfrac{68}{49}=\dfrac{-49}{49}+\dfrac{68}{49}=\dfrac{19}{49}\)

gấp ặ 

\(A=7+(\dfrac{7}{12}-\dfrac{1}{2}+3)-(\dfrac{1}{12}+5)\)

\(A=\dfrac{84}{12}+\dfrac{7-6+36}{12}-\dfrac{1+60}{12}\)

\(A=\dfrac{84+7-6+36-1-60}{12}=\dfrac{60}{12}=5\)

 = 7 + ( \(\dfrac{7}{12}\) - \(\dfrac{1}{2}\) + 3) - ( \(\dfrac{1}{12}\)+5)

=  7 + \(\dfrac{7}{12}\) - \(\dfrac{1}{2}\) + 3 - \(\dfrac{1}{12}\) -  5

= 5 + \(\dfrac{6}{12}\) - \(\dfrac{1}{2}\)

= 5 + \(\dfrac{1}{2}\) - \(\dfrac{1}{2}\)

= 5

c) \(1+2+3+4+...+97+98+99+100\)

Số số hạng: \(\left(100-1\right):1+1=100\) (số)

Tổng dãy số: \(\dfrac{\left(100+1\right)\cdot100}{2}=5050\)

d) \(2+4+6+8+...+96+98+100\)

Số số hạng: \(\left(100-2\right):2+1=50\) (số)

Tổng dãy số: \(\dfrac{\left(100+2\right)\cdot50}{2}=2550\)

e) \(15+17+19+21+...+73+75+77\)

Số số hạng: \(\left(77-15\right):2+1=32\) (số)

Tổng dãy số: \(\dfrac{\left(77+15\right)\cdot32}{2}=1472\)

Khoảng cách : `1`

Số hạng :

`(100-1)/1 + 1 = 100(số-hạng)`

Tổng :

`(100+1)xx100 :2 = 5050`

_______________________

Khoảng cách : `2`

Số hạng :

`(100-2)/2 + 1 = 50(số-hạng)`

Tổng :

`(100+2)xx50:2=2550`

__________________________

Khoảng cách : `2`

Số hạng:

`(77-15)/2 + 1 = 32(số-hạng)`

Tổng:

`(77+15)xx32:2=1472`

   ( 3879 - 459 ) : 180 x 500

=    3420 : 180 x 500

=   19 x 500

=   9500

\(x+1+x+4+x+7+...+x+28=195\)

\(10x+\left(1+4+7+...+28\right)=195\)

Xét tổng \(1+4+7+...+28=195\)

Số số hạng: \(\left(28-1\right):3+1=10\)

Tổng dãy số: \(\left(28+1\right)\times10:2=145\)

Để tìm x, ta có: 

\(10x+145=195\)

\(10x=195-145\)

\(10x=50\)

\(x=5\)

ĐS: \(x=5\)

\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)

\(=\dfrac{2^{12}.3^5-\left(2^2\right)^6.3^4}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}\)

\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(=\dfrac{3^5-3^4}{3^6+3^5}\)

\(=\dfrac{3^4.\left(3-1\right)}{3^5.\left(3+1\right)}\)

\(=\dfrac{3^4.2}{3^5.4}\)

\(=\dfrac{1}{6}\)

\(\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(=\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)

\(=\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)

\(=\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+8\right)}\)

\(=\dfrac{5.\left(-6\right)}{9}\)

\(=-\dfrac{10}{3}\)

a) ↓---↓---↓---↓---↓---↓

    5   6   7    8    9   10

b) ↓-------↓--------↓

    4        5          6

a) \(A=\left\{n\inℕ|5\le n\le10\right\}=\left\{5;6;7;8;9;10\right\}\)

Biểu diễn: -----|-----|-----|-----|-----|-----|----->

                      5     6     7     8     9    10

b) \(A=\left\{n\inℕ|3< n\le6\right\}=\left\{4;5;6\right\}\)

Biểu diễn: -----|-----|-----|----->

                      4     5     6