45/28

-Ta có: \(\dfrac{1}{2}=\dfrac{10}{20}=\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\) (có 10 số \(\dfrac{1}{20}\)).

Mà \(\dfrac{1}{20}< \dfrac{1}{19};\dfrac{1}{20}< \dfrac{1}{18};...;\dfrac{1}{20}< \dfrac{1}{11}\)

\(\Rightarrow\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}< \dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{11}\)

\(\Rightarrow\dfrac{1}{2}< S\)

a) = 1/10 - 1/11 + 1/11 -1/12 + 1/12 - 1/13 +1/13 1/14 +...+ 1/78 - 1/79

= 1/10 - 1/79

= máy tính ok

mấy câu khác bn làm tương tự là đc nhưng nhớ nhanh thêm khoảng cách giữa các mẫu nha

a)\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)

b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)=4\left(\frac{1}{7}-\frac{1}{135}\right)=4.\frac{128}{945}=\frac{456}{945}\)

c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)=4\left(\frac{1}{8}-\frac{1}{506}\right)=\frac{249}{506}\)

d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{394}\right)=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)

e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)

g) Sửa đề\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{820}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1640}\right)=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{40.41}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{40}-\frac{1}{41}\right)=2\left(1-\frac{1}{41}\right)=2.\frac{40}{41}=\frac{80}{41}\)

Bài làm:

a) \(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}\)

\(=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)

b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}\)

\(=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{135}\right)\)

\(=4.\frac{128}{945}=\frac{512}{945}\)

c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}\)

\(=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{506}\right)\)

\(=4.\frac{249}{2024}=\frac{249}{506}\)

d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}\)

\(=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{394}\right)\)

\(=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)

e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}\)

\(=\frac{4}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)\)

\(=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)

g) Phải sửa \(\frac{1}{802}\)  thành \(\frac{1}{820}\) nhé

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{820}\)

\(=1+\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+...+\frac{1}{41.20}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}\right)\)

\(=2\left(1-\frac{1}{41}\right)\)

\(=2.\frac{40}{41}=\frac{80}{41}\)

14, \(\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\)

\(=\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{\left(\sqrt{x}+2\right)\left(5\sqrt{x}-1\right)}\)

\(=\frac{\left(-7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)+\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{-7x-14\sqrt{x}+7\sqrt{x}+14+10x-2\sqrt{x}-10\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3x+20\sqrt{x}+28}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(3\sqrt{x}+14\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3\sqrt{x}+14}{5\sqrt{x}-1}\)

thank

Ta có: \(A=1-2+3-4+5-6+7-8+9\)

\(=(1+9)-(2+8)+(3+7)-(4+6)+5\)

\(=10-10+10-10+5\)

\(=5\)

Vậy  \(A=5\) 

B = 12 - 14 + 16 - 18 + ... + 2008 - 2010

B = -2 + (-2)+ (-2)+ (-2) + ...+ (-2)

B = -2 . 100

B = -200

Bạn muốn kết bạn chứ

Dấu này * là dấu nhân

Một năm rồi không có ai trả lời à 

THấy cx thương nhưng mk nhìn cái đề thì dài thật cx khó có ai có thời gian mà giải

Bạn viết như này ai hiểu

Viết lại đề lại sách nháp đi

\(\Rightarrow\frac{\left(\frac{518}{19}-\frac{342}{13}\right).\left(\frac{253}{236}-\frac{3}{318}\right)}{\left(\frac{3}{4}+x\right)\frac{27}{33}}\)=\(\frac{\frac{1}{3}.\frac{3}{36}+\frac{1}{3}.\frac{3}{14.17}}{\frac{1}{2}.\frac{2}{13.15}+\frac{1}{2}.\frac{2}{14.16}}\)

\(\frac{\frac{236}{2247}.\frac{247}{236}}{\left(\frac{3}{4}+x\right)\frac{27}{33}}\)=\(\frac{\frac{1}{3}.\left(\frac{1}{13}-\frac{1}{16}+\frac{1}{14}-\frac{1}{17}\right)}{\frac{1}{2}\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{16}-\frac{1}{17}\right)}\)

\(\frac{1}{\left(\frac{3}{4}+x\right)\frac{27}{33}}\)=\(\frac{\frac{1}{3}}{\frac{1}{2}}\)

đến đây tự làm nhé cứ ra 13/12 là được

^{_{\(\left(1-\dfrac{1}{5}\right)\times\left(1-\dfrac{1}{6}\right)\times\left(1-\dfrac{1}{7}\right)\times\left(1-\dfrac{1}{8}\right)\times\left(1-\dfrac{1}{9}\right)\)}}

^{_{\(=\dfrac{4}{5}\times\dfrac{5}{6}\times\dfrac{6}{7}\times\dfrac{7}{8}\times\dfrac{8}{9}\)}}

^{_{\(=\dfrac{4}{9}\)}}

tính nhanh đúng ko?

= 4/5 x 5/6 x 6/7 x 7/8 x 8/9

= 4 x 5 x 6 x 7 x 8/ 5 x 6 x 7 x 8 x 9

= 4/9

= 4/5 x 5/6 x 6/7 x 7/8 x 8/9

= 4/9