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technoblade
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Hang Nguyen
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OH-YEAH^^
22 tháng 7 2021 lúc 21:00

Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

Nguyễn Lê Phước Thịnh
22 tháng 7 2021 lúc 21:03

Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

Vân Nguyễn Thị
22 tháng 7 2021 lúc 21:06

Câu 1:

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(\dfrac{1}{3}-\dfrac{1}{101}\)

\(\dfrac{98}{303}\)

Câu 2 làm tương tự ở câu 1 nhé

quy pham
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Nguyễn Đức Minh
11 tháng 5 2022 lúc 14:24

ơi

Nguyễn Đức Minh
11 tháng 5 2022 lúc 14:24

Nguyễn Đức Minh
11 tháng 5 2022 lúc 14:24

không

Nguyễn Thùy Phương
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ka nekk
11 tháng 4 2022 lúc 11:07

T=4/1 . 4/3 + 4/3 . 4/5 + ... + 4/99 . 4/100

T=4/1 - 4/3 + 4/3 - 4/5 + ... + 4/99 - 4/100

T=4/1 - 4/100

T=99/25

Quỳnhh-34- 6.5 Phạm như
11 tháng 4 2022 lúc 11:11

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Nguyễn Gia Hân
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Nguyễn Kiến Hoàng
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Ta Chia Tay Đi
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Trang
9 tháng 10 2017 lúc 18:43

a) \(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\) \(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}-2.\dfrac{3}{16}=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}-\dfrac{3}{8}=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}=\dfrac{5}{8}+\dfrac{3}{8}\\ \Rightarrow\dfrac{x}{2008}=1\\ \Rightarrow x=2008\)

b) \(\dfrac{7}{x}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}\right)=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{29}{45}-\dfrac{8}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{7}{15}\\ \Rightarrow x=15\)

c) \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{15}{93}\)

\(\Rightarrow2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}\right)=\dfrac{15}{93}.2\)

\(\Rightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{30}{93}\\ \Rightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\\ \Rightarrow\dfrac{2x}{3\left(2x+3\right)}=\dfrac{10}{31}\\ \Rightarrow\dfrac{10.3\left(2x+3\right)}{31}=2x\\ \Rightarrow\dfrac{30\left(2x+3\right)}{31}=2x\\ \Rightarrow x=\dfrac{30\left(2x+3\right)}{31}:2\\ \Rightarrow x=\dfrac{30\left(2x+3\right)}{62}\\ \Rightarrow x=\dfrac{15\left(2x+3\right)}{31}\\\Rightarrow\dfrac{15\left(2x+3\right)}{x}=31\\ \Rightarrow\dfrac{30x+45}{x}=31\\ \Rightarrow30+\dfrac{45}{x}=31\\ \Rightarrow \dfrac{45}{x}=1\\ \Rightarrow x=45\)

Nguyễn Thanh Hằng
9 tháng 10 2017 lúc 12:05

a/ \(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-............-\dfrac{1}{120}=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-\left(\dfrac{1}{10}+\dfrac{1}{15}+.......+\dfrac{1}{120}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+.......+\dfrac{2}{240}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-\dfrac{3}{16}=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}=\dfrac{13}{16}\)

\(\Leftrightarrow x=1631,5\)

Vậy ..................

Vũ Ngọc Diệp
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Van Toan
3 tháng 3 2023 lúc 19:19

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)

⭐Hannie⭐
3 tháng 3 2023 lúc 19:20

`4/1.3+4/3.5+4/5.7+...+4/99.101`

`=2(2/1.3+2/3.5+2/5.7+...+2/99.101)`

`=2(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)`

`=2(1-1/101)`

`=2. 100/101`

`=200/101`

Vũ Thị Vân Anh
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Nguyễn Thanh Hằng
11 tháng 3 2017 lúc 20:52

Ta có :

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+............+\dfrac{4}{59.61}\)

\(\dfrac{A}{2}=\dfrac{2}{5.7}+\dfrac{2}{7.9}+..............+\dfrac{2}{59.61}\)

\(\dfrac{A}{2}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.......+\dfrac{1}{59}-\dfrac{1}{61}\)

\(\dfrac{A}{2}=\dfrac{1}{5}-\dfrac{1}{61}\)

\(\dfrac{A}{2}=\dfrac{56}{305}\)

\(\Rightarrow A=\dfrac{112}{305}\)

Chúc bn học tốt!!

Edogawa Conan
11 tháng 3 2017 lúc 20:52

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)

\(A=2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)

\(A=2\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(A=2\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

\(A=2.\dfrac{56}{305}\)

\(A=\dfrac{112}{305}\)

Lê Yên Hạnh
11 tháng 3 2017 lúc 20:57

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}..........+\dfrac{4}{59.61}\)

\(\dfrac{1}{2}A=\dfrac{2}{5.7}+\dfrac{2}{7.9}+.........+\dfrac{2}{59.61}\)

\(\dfrac{1}{2}A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.....+\dfrac{1}{59}-\dfrac{1}{61}\)

\(\dfrac{1}{2}A=\dfrac{1}{5}-\dfrac{1}{61}\)

\(\dfrac{1}{2}A=\dfrac{56}{305}\)

\(A=\dfrac{112}{305}\)