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

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

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

giúp giùm mình đi

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

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

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

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

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

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

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

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

\(\text{Quay lại bài toán,ta có :}\)

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

bó tay luôn

#)Giải ;

b) Đặt \(N=1+2+2^2+2^3+...+2^{2012}\)

\(\Rightarrow2N=2+2^2+2^3+2^4+...+2^{2013}\)

\(\Rightarrow2N-N=N=\left(2+2^2+2^3+2^4+...+2^{2013}\right)-\left(1+2+2^2+2^3+...+2^{2012}\right)\)

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

Thay N vào M, ta có :

\(M=\frac{2^{2013}-1}{2^{2014}-2}\)

Thêm Cho pen 

\(M=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)

Phải tính  hết nhé

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

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

\(A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3}{9}\)

\(A=\frac{1105}{48}\cdot\frac{1}{3}=\frac{1105}{144}\)

Vậy A = \(\frac{1105}{144}\)

b, Đặt A = 1 + 2 + 2^2 + 2^3 + ... + 2^2012 

2A = 2(1 + 2 + 2^2 + 2^3 + ... + 2^2012) 

2A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^2012 

2A - A = (2 + 2^2 + 2^3 + 2^4 + ... + 2^2012) - (1 + 2 + 2^2 + 2^3 + ... + 2^2012) 

A = 2^2012 - 1 

=> M = \(\frac{2^{2012}-1}{2^{2014}-2}=\frac{2^{2012}-1}{2^{2014}-2}=\frac{2^{2012}-1}{2.2.\left(2^{2012}-1\right).}=0,25\)

Vậy M = 0,25 

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

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

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

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

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

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 \(M=\frac{1}{2}\)

\(A=\frac{24.47-23}{24+27-23}.\frac{9-\frac{9}{7}+\frac{9}{11}+\frac{9}{1001}-\frac{9}{11}}{\frac{2}{1001}-\frac{2}{13}-\frac{2}{7}+\frac{2}{11}+2}\)

co sai de ko bn

\(\Rightarrow\)

Mình nhầm đề câu A tí nha !

\(A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{9-\frac{9}{7}+\frac{9}{11}+\frac{9}{1001}-\frac{9}{13}}{\frac{2}{1001}-\frac{2}{13}-\frac{2}{7}+\frac{2}{11}+2}\)

Ai làm nhanh 3 k !

cau A nha:

\(\Rightarrow\frac{1105}{48}.\frac{9.\left(1-\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{2.\left(\frac{1}{1001}-\frac{1}{13}-\frac{1}{7}+\frac{1}{11}+1\right)}\)

\(\Rightarrow\frac{1105}{48}.\frac{9}{2}\)

\(\Rightarrow\frac{1105.9}{48.2}\)

???