1+2+3+4+5+6+7+8+9+10+11+12+13-1-2-3-4-5-6-7-8-9-10-11-12

=(1-1)+(2-2)+(3-3)+(4-4)+(5-5)+(6-6)+(7-7)+(8-8)+(9-9)+(10-10)+(11-11)+(12-12)+13

=  0  +  0  +  0  +  0  +   0  +  0  +  0  +  0  +  0  + 0  + 0   +  0  +  13

=                                 13

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 - 1 - 2 - 3 - 4 - 5 - 6 - 7- 8 - 9 - 10 

= (1+12) + (2+11) + (3+10) + ( 4+9) + (5+8) + (6+7) + 13 -  (1+11) - ( 2+10) - ( 3+9) - (4+8) - (5+7) + 6 + 12

= 13 * 7 - 12 * 6 + 6

= 13

Ai k mk mk k lại

= 13 nhé hihi nha

coi lại câu a

1.1+1.1+1.1+1.1+1.1/1.1+1.1+9.3+6.8+.2

Lời giải:
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}\)

\(=5-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)

\(=5-\left(\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}\right)\)

\(=5-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)

\(=5-\left(1-\frac{1}{6}\right)=5-\frac{5}{6}=\frac{25}{6}\)

a) 5 – 4 – 3 – 2 – 1 + 11 + 12 + 13 + 14 + 15

= (5 + 15) + (14 – 4) + (13 – 3) + (12 – 2) + (11 – 1)

= 20 + 10 + 10 + 10 + 10

= 60 

b) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 +11

= (1 + 3 + 5 + 7 + 9 + 11) – 2 – 4 – 8 – 10

= 36 – 24

= 12

a) 5 – 4 – 3 – 2 – 1 + 11 + 12 + 13 + 14 + 15

= (5 + 15) + (14 – 4) + (13 – 3) + (12 – 2) + (11 – 1)

= 20 + 10 + 10 + 10 + 10

= 60 

b) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 +11

= (1 + 3 + 5 + 7 + 9 + 11) – 2 – 4 – 8 – 10

= 36 – 24

= 12

P=\(\frac{\frac{44}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}=\frac{\frac{44\cdot1}{132}+\frac{-33\cdot1}{132}+\frac{60\cdot1}{132}}{\frac{55\cdot1}{132}+\frac{132\cdot1}{132}-\frac{84\cdot1}{132}}\)

\(P=\frac{\frac{1}{132}\left(44-33+60\right)}{\frac{1}{132}\left(55+132-84\right)}=\frac{\frac{1}{132}\cdot71}{\frac{1}{132}\cdot103}\)

\(P=\frac{71}{103}\)

\(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{6}{11}}=1\)

ta có:\(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}=\frac{88}{132}-\frac{33}{132}+\frac{60}{132}=\frac{88-33+60}{132}=\frac{115}{132}\)(1)

        \(\frac{5}{12}+1-\frac{6}{11}=\frac{55}{132}+\frac{132}{132}-\frac{72}{132}=\frac{115}{132}\)(2)

từ (1)(2) suy ra \(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{6}{11}}=1\)

A=(2+3+...+13)-(1+2+...+12)=2+3+...+13-1-2-...-12=(13-1)+(2-2)+(3-3)+...+(12-12)=12

12 nha cậu .

\(\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)

=         1             +                  2                +               2

= 5

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)

=     \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

=\(1-\frac{1}{7}\)

=\(\frac{6}{7}\)

a. bằng77

b. bằng 13

lớp 2 đã học cái này rồi à

a. bằng77

b. bằng 13