Mình​ cũng đang k pit lm câu này

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a,Gọi tổng trên là A.

Xét \(\frac{4}{5}-\frac{4}{7}=\frac{8}{35};...;\frac{4}{59}-\frac{4}{61}=\frac{8}{3599}\)=>\(A=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{59}-\frac{4}{61}\right)\)\(=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{61}\right)=\frac{1}{2}.\frac{224}{305}=\frac{112}{305}\)

b,Gọi tổng trên là B

Theo đề bài ta có:\(B=\frac{24.47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)=\(\frac{\left(23+1\right).47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}=\frac{47.23+24}{24+47.23}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{3.\left(3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}\right)}\)\(=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}}=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3.\left(1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{1}{3}\)

\(2\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{61}\right)=2\left(\frac{61-5}{305}\right)=2.\frac{56}{305}=\frac{112}{305}\)

Đặt A=B*C

B=\(\frac{24\cdot47-23}{24+47-23}=\frac{1128-23}{71-23}=\frac{1105}{48}\)

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

Suy ra A =\(\frac{1105}{144}\)

\(A=\frac{24.47-23}{24+47-23}.\frac{3+\frac{3}{7}+\frac{3}{11}-\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}+9}\)

\(A=\frac{1105}{48}.\frac{3.\left(1+\frac{1}{7}+\frac{1}{11}-\frac{1}{1001}+\frac{1}{13}\right)}{9.\left(\frac{1}{1001}-\frac{1}{11}+\frac{1}{7}+1\right)}\)

\(A=\frac{1105}{48}.\frac{3.\frac{1311}{1001}}{9.\frac{1054}{1001}}\)

\(A=\frac{1105}{48}.\frac{3933}{1001}:\frac{9468}{1001}\)

\(A=\frac{1105}{48}.\frac{437}{1052}\)

\(A\approx9,56\)

Chú ý : Dấu xấp xỉ \(\approx\)

10 nha anh k em nha

đề bài thiếu

BẠN XEM LẠI CÁI ĐỀ XEM ĐÚNG KO

\(\frac{11}{12}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}=\frac{11}{12}.\frac{1}{3}=\frac{11}{36}\)

Bài 1: 

\(A=\frac{24.47-22}{24+47.23}.\frac{5+\frac{5}{7}+\frac{5}{11}-\frac{5}{13}+\frac{5}{1001}}{6+\frac{6}{7}+\frac{6}{11}-\frac{6}{13}+\frac{6}{1001}}\)\(=\frac{47.23+47-22}{24.47.23}.\frac{5\left(1+\frac{1}{7}+\frac{1}{11.}-\frac{1}{13}+\frac{1}{1001}\right)}{6\left(1+\frac{1}{7}+\frac{1}{11}-\frac{1}{13}+\frac{1}{1001}\right)}\) 

                                                                         \(=\frac{47.23+24}{24+47.23}.\frac{5}{6}\) 

                                                                         \(=1.\frac{5}{6}=\frac{5}{6}\) 

Bài 2: 

\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\) 

                                      \(=3^{28}-3^{27}-3^{26}\) 

                                      \(=3^{22}\left(3^6-3^5-3^4\right)\) 

                                      \(=3^{22}.405\) chia hết cho 405 

 =>đpcm

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

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

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

giúp giùm mình đi

\(A=\frac{24.47-23}{24+47-23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\) 

\(A=\frac{1105}{28}.\)\(\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{9+\frac{9}{7}-\frac{9}{11}+\frac{9}{1001}-\frac{9}{13}}\)

\(A=\frac{1105}{28}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}\)

\(A=\frac{1105}{28}.\frac{3}{9}\)

\(A=\frac{1105}{84}\)

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

Đặt \(A=1+2+2^2+2^3+...+2^{2012}\)

Suy ra \(2.A=2+2^2+2^3+2^4+...+2^{2013}\)

Khi đó \(2.A-A=2^{2013}-1\)hay \(A=2^{2013}-1\)

Do đó : \(M=\frac{A}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{1}{2}\)

          Vậy \(M=\frac{1}{2}\)

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

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

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

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

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

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

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

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

\(\text{Quay lại bài toán,ta có :}\)

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