(A) \(\frac{2}{3} + \frac{{ - 4}}{6} = \frac{4}{6} + \frac{{ - 4}}{6} = 0\) => A sai

(B) \(\frac{2}{3}.\frac{{ - 1}}{5} = \frac{{ - 2}}{{15}}\) mà \(\frac{{3 - 2}}{5} = \frac{1}{5}\) => B sai

(C) \(\frac{2}{3} - \frac{3}{5} = \frac{{10}}{{15}} - \frac{9}{{15}} = \frac{1}{{15}}\) => C đúng

(D) \(\frac{3}{5}:\frac{3}{{ - 5}} = \frac{3}{5}.\frac{{ - 5}}{3} = \frac{{ - 15}}{{15}} =  - 1\) => D sai

=> Chọn C.

\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)

\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)

\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)

giúp giùm mình đi

\(a)A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}\)

\(=\frac{(23+1)\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}=\frac{47-23+24}{47-23+24}\cdot\frac{3(1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13})}{3(3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11})}\)

\(=\frac{1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13}}{3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11}}=\frac{1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11}}{3(1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11})}=\frac{1}{3}\)

\(b)\)\(\text{Đặt A = }1+2+2^2+2^3+...+2^{2012}\)

\(2A=2(1+2^2+2^3+...+2^{2012})\)

\(2A=2+2^2+2^3+...+2^{2013}\)

\(2A-A=(2+2^2+2^3+2^4+...+2^{2013})-(1+2+2^2+2^3+...+2^{2012})\)

\(\Rightarrow A=2^{2013}-1\)

\(\text{Quay lại bài toán,ta có :}\)

\(B=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2(2^{2013}-1)}=\frac{1}{2}\)

các bn ơi giải giúp mình đi mà

A. Đặt A= biểu thức đã cho

=>\(\frac{A}{3}=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

=>\(\frac{A}{3}.2=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

=>\(\frac{2A}{3}-\frac{A}{3}=2-\frac{1}{2^9}\)

=>\(A=\frac{3\left(2^{10}-1\right)}{2^9}\)

B. Đặt B=biểu thức đã cho

\(\Rightarrow\frac{B}{2}=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

\(=\frac{1}{3}-\frac{1}{2017}=\frac{2014}{6051}\)

\(\Rightarrow B=\frac{4028}{6051}\)

A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
  =(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)

Đặt \(A=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

Nên \(2.A=6+3+\frac{3}{2}+....+\frac{3}{2^8}\)

Suy ra \(2.A-A=6-\frac{3}{2^9}\Rightarrow A=6-\frac{3}{2^9}\)

Vậy giá trị biểu thức là : \(6-\frac{3}{2^9}\)

đặt \(A=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

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

đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)( 1 )

\(2B=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)( 2 )

Lấy ( 2 ) - ( 1 ) ta được :

\(B=2-\frac{1}{2^9}\)

\(\Rightarrow A=3.\left(2-\frac{1}{2^9}\right)\)

\(\Rightarrow A=6-\frac{3}{2^9}\)

bạn ơi bạn giải dc chưa