\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)

\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)

\(=12.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=12.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

\(=12.\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=6.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=6.\frac{8}{33}\)

\(=\frac{16}{11}\)

_Chúc bạn học tốt_

Mua ha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihi ha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihi ha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihi ha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihiha ha ha ha hihihihihihihihihihi

tại sao phải chia cho 1/2 z bn ?
 

\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\)

=\(\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\)

=\(\dfrac{16}{11}\)

Giải:

\(\dfrac{1212}{1515}+\dfrac{1212}{3535}+\dfrac{1212}{6363}+\dfrac{1212}{9999}\) 

\(=\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}\) 

\(=12.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)\) 

\(=12.\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)\) 

\(=6.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\) 

\(=6.\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\) 

\(=6.\dfrac{8}{33}\) 

\(=\dfrac{16}{11}\)

\(=\dfrac{12}{15}+\dfrac{12}{35}+\dfrac{12}{63}+\dfrac{12}{99}=12\cdot\left[\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}\right]=12\cdot\left[\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}\right]=12\cdot\dfrac{1}{2}\cdot\left[\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right]=6\cdot\left[\dfrac{1}{3}-\dfrac{1}{11}\right]=6\cdot\dfrac{8}{33}=\dfrac{48}{33}=\dfrac{16}{11}\)

a)=4.25x(58.47+41.53)-125

=4.25x100-125

=425-100

=325

= \(\frac{12}{15}\) +\(\frac{12}{35}\)+\(\frac{12}{63}\)+\(\frac{12}{99}\)

= 12 x (\(\frac{1}{15}\)+\(\frac{1}{35}\)+\(\frac{1}{63}\)+\(\frac{1}{99}\))

= 12 x ( \(\frac{1}{3x5}\)+\(\frac{1}{5x7}\)+\(\frac{1}{7x9}\)+\(\frac{1}{9x11}\))

= 12 x \(\frac{1}{2}\) x    ( \(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))

= 6 x (   \(\frac{1}{3}\) -  \(\frac{1}{11}\))

=  6 x \(\frac{8}{33}\)

= \(\frac{48}{33}\)=\(\frac{16}{11}\)

      Nhớ tk nha

lấy tất cả chia cho 1010

\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)

\(=1212.\left(\frac{1}{15.101}+\frac{1}{35.101}+\frac{1}{63.101}+\frac{1}{99.101}\right)\)

\(=12.101.\frac{1}{101}.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=6.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=6.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=6.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=6.\left(\frac{11}{33}-\frac{3}{33}\right)\)

\(=6.\frac{8}{33}\)

\(=\frac{16}{11}\)

\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)

\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)

\(=12\cdot\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=12\cdot\frac{4}{33}\)

\(=\frac{16}{11}\)

\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)

\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)

\(=12\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=12\left(\frac{1}{3\cdot5}+\frac{1}{3\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\)

\(=12\cdot\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{3\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)

\(=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=6\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=6\cdot\frac{8}{33}\)

\(=\frac{48}{33}\)

Sorry , mình làm hơi tắt 1 xíu

Ta có : \(12\cdot\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=6\cdot\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)\)

\(=6\cdot\left(\frac{2}{3\cdot5}+\cdot\cdot\cdot+\frac{2}{9\cdot11}\right)\)

\(=6\cdot\left(\frac{1}{3}-\frac{1}{5}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{11}\right)\)

\(=6\cdot\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=6\cdot\frac{8}{33}\)

\(=\frac{16}{11}\)

\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)

\(=1212.\left(\frac{1}{1515}+\frac{1}{3535}+\frac{1}{6363}+\frac{1}{9999}\right)\)

Sau đó tự tính rồi làm típ nha 

Study well 

\(1,\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}=6\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=6\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).Tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{11}\right)=6\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{48}{33}=\frac{16}{11}\)

\(2,\left(x+1\right)+\left(x+2\right)+.....+\left(x+211\right)=211x+\left(1+2+....+211\right)=211x+\frac{212.211}{2}=211x+22366=23632\Leftrightarrow211x=23632-22366=1266\Leftrightarrow x=6\)

a, \(14:\left(4\frac{2}{3}:1\frac{5}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)

=> \(14:\frac{28}{9}+14:\frac{14}{9}=>14.\frac{9}{28}+14.\frac{9}{14}\)

=> 14. ( \(\frac{9}{28}+\frac{9}{14}\) )

=> \(14.\frac{27}{28}=\frac{419}{28}\)

b, \(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)

=> \(\frac{4}{5}+\frac{12}{35}+\frac{4}{21}+\frac{4}{33}\)

=> \(\frac{8}{7}+\frac{24}{77}=\frac{16}{11}\)

bài 2 :

( x + 1 ) + ( x + 2 ) + ... + ( x + 211 ) = 23632

=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 211 ) = 23632

=> 211x + 22366 = 23632

=> 211x = 23632 - 22366

=> 211x = 1266

=> x = 1266 : 211

x = 6

bước 1: rút về số tối giản

bước 2: tìm MSC

bước 3: tính...

\(\frac{1212}{1515}:\frac{2727}{2525}\)

Rút gọn : \(\frac{1212}{1515}=\frac{4}{5}\)

\(\frac{2727}{2525}=\frac{27}{25}\)

\(=>\frac{4}{25}.\frac{25}{27}=\frac{20}{27}\)

~ chúc bn hok tốt ~

cám ơn nha

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