a, \(3332:\left(4.49\right)=3332:196=17\)

b, \(72.3.125=\left(72.125\right).3=9000.3=27000\)

c, \(123:1,3+123:7,8+123:0,9=123:\left(1,3+7,8+0,9\right)=123.10=1230\)

d, \(15.24+15.57+81.29+81.56=15\left(24+57\right)+81\left(29+56\right)=15.81+81.85=81\left(15+85\right)=81.100=8100\)

\(\dfrac{6}{5.7}+\dfrac{6}{7.9}+...+\dfrac{6}{59.61}\)

\(=3\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)

\(=3\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(=3\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

\(=\dfrac{3.56}{305}\\ =\dfrac{168}{305}\)

\(=3\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)=3\cdot\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=3\cdot\dfrac{56}{305}=\dfrac{168}{305}\)

Giải:

\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\) 

\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\) 

\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\) 

\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)

\(B=\dfrac{3}{2}\times\dfrac{47}{150}\) 

\(B=\dfrac{47}{100}\) 

Chúc em học tốt!

S=1/5-1/7+1/7-1/9+1/9-1/11+...+1/93-1/95

S=1/5-1/95

S=18/95

sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

trả lời hiền thương đề bài của bạn ấy là đúm gòi nha

Đặt :

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(\Rightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(\Rightarrow2A=\frac{1}{3}-\frac{1}{11}\)

\(\Rightarrow2A=\frac{3}{11}\)

\(\Rightarrow A=\frac{3}{22}\)

1/3 x 5 = 1/3 - 1/5 ;  1/5 x 7 = 1/5 - 1/7 ; làm  tương tự  với số còn lại

Ta được :

1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

= 1/3 - 1/11

=

= 11/33 - 3/33

= 8/33

\(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{19\times21}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)

                                                    \(=1-\frac{1}{21}=\frac{20}{21}\)

đúng cái nhé

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

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

\(=\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)