a: =110-200=-90

a: =110-200=-90

b: =25+50-1=74

c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)

\(=\dfrac{255}{103}\)

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

câu 45 nhân 11 cách giải làm sao

a) \(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{97.99}\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{3}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{3}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\frac{3}{2}.\frac{32}{99}\)

\(=\frac{16}{33}\)

b)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)

\(=1-\frac{1}{103}\)

\(=\frac{102}{103}\)

này Nguyễn Thanh Tùng có chắc chắn ko vậy ?

Tính nhanh:

a) 1-2-3+4+5-6-7+8+9-10-.......+200+201-202-203

Đặt \(A=\text{1-2-3+4+5-6-7+8+9-10-.......+200+201-202-203}\)

\(A=1+\left(2-3-4+5\right)+\left(6-7-8=9\right)+\left(10-11-12+13\right)+... \)\(+\left(298-299-300+301\right)+302\)

\(A=1+0+0+0+...+0+302\)

\(A=1+302\)

\(A=303\)

Phần B làm sau nha!