= \(\frac{9}{1x4}+\frac{9}{4x7}+\frac{9}{7x10}+.........+\frac{9}{19x22}+\frac{9}{22x25}\)

= \(\frac{1}{3}x\left(\frac{9}{1}-\frac{9}{4}\right)+\left(\frac{9}{4}-\frac{9}{7}\right)x\frac{1}{3}+........+\left(\frac{9}{22}-\frac{9}{25}\right)x\frac{1}{3}\)

= \(\frac{1}{3}\left(\frac{9}{1}-\frac{9}{4}+\frac{9}{4}-\frac{9}{7}+....+\frac{9}{22}-\frac{9}{25}\right)\)

= \(\frac{1}{3}x\left(\frac{9}{1}-\frac{9}{25}\right)\)

= \(\frac{1}{3}x\frac{216}{25}\)

= \(\frac{72}{25}\)

nhớ ********** nha bn thân

\(\frac{9}{4}+\frac{9}{28}+\frac{9}{70}+\frac{9}{130}+...+\frac{9}{418}+\frac{9}{550}\)

\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{19.22}+\frac{3}{22.25}\right)\)

\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{22}-\frac{1}{25}\right)\)

\(=3\left(1-\frac{1}{25}\right)\)

\(=3.\frac{24}{25}=\frac{72}{25}\)

Có: \(\frac{9}{10!}=\frac{9}{10!}\)

\(\frac{9}{11!}< \frac{10}{11!}=\frac{11-1}{11!}=\frac{11}{11!}-\frac{1}{11!}=\frac{1}{10!}-\frac{1}{11!}\)

\(\frac{9}{12!}< \frac{11}{12!}=\frac{12-1}{12!}=\frac{12}{12!}-\frac{1}{12!}=\frac{1}{11!}-\frac{1}{12!}\)

............

\(\frac{9}{1000!}< \frac{999}{1000!}=\frac{1000-1}{1000!}=\frac{1000}{1000!}-\frac{1}{1000!}=\frac{1}{999!}-\frac{1}{1000!}\)

\(\Rightarrow\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{1}{1000!}< \frac{9}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{999!}-\frac{1}{1000!}\)

\(\Rightarrow\frac{9}{10!}+\frac{9}{11!}+...+\frac{1}{1000!}< \frac{10}{10!}-\frac{1}{1000!}=\frac{1}{9!}-\frac{1}{1000!}< \frac{1}{9!}\)

\(\Rightarrow\frac{9}{10!}+\frac{9}{11!}+...+\frac{9}{1000!}< \frac{1}{9!}\)

\(\Rightarrowđpcm\)

đặt tên là B

B =910!+911!+912!+.............+91000!B=910!+911!+912!+.............+91000!

Ta thấy :

910!=10−110!=19!−110!910!=10−110!=19!−110!

911!<11−111!=110!−111!911!<11−111!=110!−111!

91000!<1000−11000!=1999!−11000!91000!<1000−11000!=1999!−11000!

⇒B<19!−110!+110!−111!+...
.....+1999!−11000!⇒B<19!−110!+110!−111!+............+1999!−11000!

B<19!−11000!B<19!−11000!

⇒B<19!→đpcm

dùm e đi ạ :v

đề sai e ưi

sửa ik e

\(A=9\left(\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+...+\frac{1}{34x35}+\frac{1}{35.36}\right)\)

\(A=9\left(\frac{5-4}{4x5}+\frac{6-5}{5x6}+\frac{7-6}{6x7}+...+\frac{35-34}{34x35}+\frac{36-35}{35x36}\right)\)

\(A=9\left(\frac{1}{4}-\frac{1}{36}\right)=9x\frac{8}{36}=2\)

Toán lớp 5 à ?

Sao khó quá vậy ?

Nhìn là ngán chả mún làm ! phù...............

Mí bạn k nha !

chung minh thu ha ban

Đặt A = 9/4  + 9/28 +.. + 9/550

       A = 9/1.4 + 9/4.7 +... + 9/22.25

       A = 3( 3/1.4 + 3/4.7 + .. + 3/22.25)

        A  = 3 . (1/1 - 1/4 + 1/4  - 1/7 + ... +1/22 - 1/25)

         A = 3 (1 - 1/25)

       A   = 3. 24 / 25

       A   =  72/25

chỗ giữa 2 phân số là dấu nhân hay sao mà chả thấy dấu j thế?

Trà My:là dấu nhân thế cx hỏi

Ta có:

\(D = \frac{{\sin \frac{{7\pi }}{9} + \sin \frac{\pi }{9}}}{{\cos \frac{{7\pi }}{9} - \cos \frac{\pi }{9}}} = \frac{{2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\cos \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}}{{ - 2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\sin \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}} = -\cot \frac{\pi }{3} = -\frac{{\sqrt 3 }}{3}\)

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9.\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(A=9.\frac{99}{100}\)

\(A=\frac{891}{100}\)

kết quả là 891/100 nha

\(A=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9.\left(1+\left[-\frac{1}{2}+\frac{1}{2}\right]+\left[-\frac{1}{3}+\frac{1}{3}\right]+...+\left[-\frac{1}{99}+\frac{1}{99}\right]-\frac{1}{100}\right)\)

\(A=9.\left(1+0+0+...+0-\frac{1}{100}\right)\)

\(A=9.\left(1-\frac{1}{100}\right)\)

\(A=9.\left(\frac{100}{100}-\frac{1}{100}\right)=9.\left(\frac{99}{100}\right)\)

\(A=\frac{891}{100}=8\frac{91}{100}\)

k cho mk nha

\(A=\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(A=\frac{9.1}{1.2.1}+\frac{9.1}{2.3.1}+...+\frac{9.1}{98.99.1}+\frac{9.1}{99.100.1}\)

\(A=1\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=1\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=1\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(A=1.\frac{99}{100}\)

\(A=\frac{99}{100}\)