Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}\)
\(=\frac{(23+1)\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}=\frac{47-23+24}{47-23+24}\cdot\frac{3(1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13})}{3(3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11})}\)
\(=\frac{1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13}}{3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11}}=\frac{1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11}}{3(1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11})}=\frac{1}{3}\)
\(b)\)\(\text{Đặt A = }1+2+2^2+2^3+...+2^{2012}\)
\(2A=2(1+2^2+2^3+...+2^{2012})\)
\(2A=2+2^2+2^3+...+2^{2013}\)
\(2A-A=(2+2^2+2^3+2^4+...+2^{2013})-(1+2+2^2+2^3+...+2^{2012})\)
\(\Rightarrow A=2^{2013}-1\)
\(\text{Quay lại bài toán,ta có :}\)
\(B=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2(2^{2013}-1)}=\frac{1}{2}\)
a = 3/54 + 5/126 + 7/294 + 8/609 + 13/1218
=> a = 3/6.9 + 5/9.14 + 7/14.21 + 8/21.29 + 13/29.42
=> a = 1/6 - 1/9 + 1/9 - 1/14 + 1/14 - 1/21 + 1/21 - 1/29 + 1/29 - 1/42
=> a = 1/6 - 1/42
=> a = 7/42 - 1/42
=> a = 6/42 = 1/7
\(A=\frac{24.47-23}{24+47-23}.\frac{3+\frac{3}{7}+\frac{3}{11}-\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}+9}\)
\(A=\frac{1105}{48}.\frac{3.\left(1+\frac{1}{7}+\frac{1}{11}-\frac{1}{1001}+\frac{1}{13}\right)}{9.\left(\frac{1}{1001}-\frac{1}{11}+\frac{1}{7}+1\right)}\)
\(A=\frac{1105}{48}.\frac{3.\frac{1311}{1001}}{9.\frac{1054}{1001}}\)
\(A=\frac{1105}{48}.\frac{3933}{1001}:\frac{9468}{1001}\)
\(A=\frac{1105}{48}.\frac{437}{1052}\)
\(A\approx9,56\)
Chú ý : Dấu xấp xỉ \(\approx\)
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
Đề bài yêu cầu tính
cái dấu chấm ở câu đầu là dấu gì vậy