BẠN XEM LẠI CÁI ĐỀ XEM ĐÚNG KO

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

A=(24.47-23)/(24+47-23) . [3(1+1/7-1/11-1/13+1/1001)]/[9(1+1/7-1/11-1/13+1/1001)]

=1105/48 . 3/9 =1105/144

Mình​ cũng đang k pit lm câu này

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

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

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

giúp giùm mình đi

#)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