C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.

chắc là 51 nhé!!!!!!!!!!!!!!!!!!!!!!!!!

\(1-2+3-4+5-6+.......+97-98+99-100+101\)

\(=\left(1-2\right)+\left(3-4\right)+\left(4-5\right)+.....+\left(97-98\right)+\left(99-100\right)+101\)

\(=50.\left(-1\right)+101=51\)

D = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.

A= 1+1+1+1+1..........+1

A có số số 1 là 

(100-2):2 +1= 50

tổng đó là

(100+2).50:2=2550

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

100+99-98-97+96+95-94-93+....+4+3-2-1

=(100+99-98-97)+(96+95-94-93)+....+(4+3-2-1)

=4+4+....+4(25 số hạng)

=4.25=100

  100 + 99  - 98 - 97 + ......... + 4 + 3 - 2 -1

= ( 100 + 99 - 98 - 97 ) + ........+ ( 4 + 3 - 2 -1 ) ( 25 nhóm )

= 4 + 4 + ....... + 4 ( 25 số 4 )

= 4 .25

 = 100

tui lớp 5 nhưng khó v~ chóe

a = 1 + 1 + ... +1 +2

a = 49 + 2

a = 51

ai thấy tớ đúng thì ủng hộ nha

Tớ đoán chắc là toán lớp 6

Giải

A= (100-99)+(98-97)+...+(4-3)+2

A=-1+(-1)+...+(-1)+2( 49 số -1)

A=-1.49+2

A= -49+2=-47

Vậy A= -47

A

100-99+98-97+...+4-3+2 = 1 x 50 = 50 

cảm ơn bạn nhưng bạn sai rồi đáp án đúng là :100-99+98-97+...+4-3+2

                                                                  =(100-99)+(98-97)+...+(4-3)+2

                                                                  =1         +   1     +...+    1  +2 (có 49 số 1)

                                                                  =49+2=51 

100 - 99 + 98 - 97 + ... + 4 - 3 + 2 ( có 99 số )

= 100 - 99 + 98 - 97 + ... + 4 - 3 + 2 ( có 49 cặp ) 

= 1 + 1 + ....... + 1 + 2 ( có 49 số 1 )

= 1 x 49 + 2

= 49 + 2

= 51

chắc chắn rằng 51 đó cái thằng quỉ sứ