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\(a)A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}\)
\(=\frac{(23+1)\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}=\frac{47-23+24}{47-23+24}\cdot\frac{3(1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13})}{3(3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11})}\)
\(=\frac{1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13}}{3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11}}=\frac{1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11}}{3(1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11})}=\frac{1}{3}\)
\(b)\)\(\text{Đặt A = }1+2+2^2+2^3+...+2^{2012}\)
\(2A=2(1+2^2+2^3+...+2^{2012})\)
\(2A=2+2^2+2^3+...+2^{2013}\)
\(2A-A=(2+2^2+2^3+2^4+...+2^{2013})-(1+2+2^2+2^3+...+2^{2012})\)
\(\Rightarrow A=2^{2013}-1\)
\(\text{Quay lại bài toán,ta có :}\)
\(B=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2(2^{2013}-1)}=\frac{1}{2}\)
\(A=\frac{24.47-23}{24+47-23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(A=\frac{1105}{28}.\)\(\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{9+\frac{9}{7}-\frac{9}{11}+\frac{9}{1001}-\frac{9}{13}}\)
\(A=\frac{1105}{28}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}\)
\(A=\frac{1105}{28}.\frac{3}{9}\)
\(A=\frac{1105}{84}\)
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 \(M=\frac{1}{2}\)
#)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
Love you sai phần b rồi
\(2A=2+2^2+2^3+...+2^{2013}\)nhé
#) Tks CK
kia ghi 2^2012 nhé bn :>
à uk sai :)))) quên chưa nhân . vs 2
#)Khi gấp lên 2 lần thì số mũ sẽ tăng lên nhé bạn !
từ 2^2012 thành 2^2013 vì là số 2 cùng nhau và gấp lên 2 lần thì tích 2.2.2.....2 cũng sẽ được thêm một số 2 nên tích thành 2.2.2.....2 = 2^2013