Lời giải:
$A=1+2-3+4-5+6-7+8-9+...+102-103+104$

$=1+(2+4+6+8+...+102+104)-(3+5+7+...+103)$

Tính tổng $2+4+6+...+104$

Số số hạng: $(104-2):2+1=52$

$2+4+6+...+104=(104+2)\times 52:2=2756$

Tính tổng $3+5+7+...+103$

Số số hạng: $(103-3):2+1=51$

$3+5+7+...+103=(103+3)\times 51:2=2703$

$A=1+2756-2703=2757-2703=54$

1 + 2 - 3 - 4 + 5 + 6 -7 -8 + 9+...+101 + 102 - 103 - 104

= (1+2 - 3 - 4) + (5 + 6 -7 - 8) +...+ (101 + 102 - 103 - 104)

= -4.24

= -96

\(1.\) \(\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{7+32-24}{36}=\dfrac{5}{12}.\)

\(2.\) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}=\dfrac{-9-2-7}{18}+\dfrac{4+3}{7}=\dfrac{-18}{18}+\dfrac{7}{7}=-1+1=0.\)

\(3.\) \(-\dfrac{10}{3}+\dfrac{13}{10}-\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{13+1}{10}+\dfrac{-20-1}{6}=\dfrac{14}{10}+\dfrac{-21}{6}=\dfrac{7}{5}-\dfrac{7}{2}=-\dfrac{21}{10}.\)

\(4.\) \(\dfrac{10}{17}-\dfrac{5}{13}-\left(-\dfrac{7}{17}\right)-\dfrac{8}{13}+\dfrac{11}{25}=\dfrac{10+7}{17}+\dfrac{-5-8}{13}+\dfrac{11}{25}=\dfrac{17}{17}-\dfrac{13}{13}+\dfrac{11}{25}=\dfrac{11}{25}.\)

1: \(=\dfrac{7}{36}+\dfrac{32}{36}+\dfrac{-24}{36}=\dfrac{15}{36}=\dfrac{5}{12}\)

2: \(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\dfrac{3}{7}+\dfrac{4}{7}\)

\(=\dfrac{-9-2-7}{18}+1=-1+1=0\)

3: \(=\left(-\dfrac{10}{3}-\dfrac{1}{6}\right)+\dfrac{14}{10}\)

\(=-\dfrac{21}{6}+\dfrac{14}{10}=\dfrac{-7}{2}+\dfrac{14}{10}=\dfrac{-35+14}{10}=-\dfrac{21}{10}\)

4: \(=\left(\dfrac{10}{17}+\dfrac{7}{17}\right)-\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\dfrac{11}{25}=\dfrac{11}{25}\)

câu 45 nhân 11 cách giải làm sao

\(\text{1+6+9+8+5+965+9+98+8+595+986+8+8+98+986+9}\)

\(=3799\)

học tốt nhé!!

@@

TL 

= 3799 

~HT~ 

tao lớp 4 ok

Ko hiểu đề bài!

đầu tiên ta tìm số hạng của dãy :

       ( 110 - 1 ) : 1 + 1 = 110 ( số )

tiếp ta tìm tổng :

       ( 110 + 1 ) x 110 : 2 = 6105 

 ĐS : 6105

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