Chứng minh rằng
\(\frac{1.3.5.7....39}{21.22.23....40}=\frac{1}{2^{20}}\)
Chứng minh rằng : \(\dfrac{1.3.5.7...39}{21.22.23..40}=\dfrac{1}{2^{20}}\)
CM: \(\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\dfrac{1}{2^{20}}\)
Biến đổi vế trái:
\(\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{22\cdot24\cdot26\cdot\cdot\cdot40}\)
\(=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{2\cdot11\cdot2^3\cdot3\cdot2\cdot13\cdot2^2\cdot7\cdot2\cdot15\cdot2^5\cdot2\cdot17\cdot2^2\cdot9\cdot2\cdot19\cdot2^3\cdot5}\)
\(=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{\left(3\cdot5\cdot7\cdot\cdot\cdot19\right)2^{20}}\)
\(=\dfrac{1}{2^{20}}\)
so sánh \(U=\frac{1.3.5.7....39}{21.22.23...40}\)và \(V=\frac{1}{2^{20}-1}\)
so sánh \(U=\frac{1.3.5.7....39}{21.22.23...40}\)và \(V=\frac{1}{2^{20}-1}\)
Chứng minh rằng \(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
Nhân cả tử và mẫu của phân số \(\frac{1.3.5...39}{21.22.23...40}\) ta được:
\(\frac{\left(1.3.5...39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}=\frac{1.2.3...39.40}{21.22.23...40.\left[\left(1.2\right).\left(2.2\right)....\left(2.20\right)\right]}\)
\(=\frac{1.2.3...39.40}{21.22.23...40.\left(1.2.3...20\right).2^{30}}=\frac{1.2.3...39.40}{1.2.3...20.21....40.2^{20}}=\frac{1}{2^{20}}\)
Suy ra điều phải chứng minh.
Chứng minh rằng: \(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
Nhân cả tử và mẫu với 2.4.6.....40, ta được:
\(\frac{1.3.5.....39}{21.22.23.....40}=\frac{\left(1.3.5.....39\right)\left(2.4.6.....40\right)}{\left(21.22.23.....40\right)\left(1.2.3.....20\right).2^{20}}=\frac{1}{2^{20}}\left(đpcm\right)\)
Vậy \(\frac{1.3.5.....39}{21.22.23.....40}\)=\(\frac{1}{2^{20}}\)
Chứng minh rằng: \(\frac{1.3.5.....39}{21.22.23.....40}=\frac{1}{2^{20}}\)
Nhân cả từ và mẫu với 2 . 4 . 6 ... 40 ta được:
\(\frac{1.3.5...39}{21.22.23...40}=\frac{\left(1.3.5...39\right)\left(2.4.6...40\right)}{\left(21.22.23...40\right)\left(2.4.6...40\right)}=\frac{1.2.3.4...39.40}{21.22.23...40.\left(1.2.3...20\right).2^{20}}=\frac{1}{2^{20}}\)(đpcm)
Vậy \(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
Chứng minh rằng \(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2}\)
Ta có:\(\frac{1.3.5......39}{21.22.23........4}=\frac{1.3.5....39.2.4.6...40}{21.22.23......40.2.4.6.....40}\)
=\(\frac{40!}{21.22....40\left(1.2.3....20\right).2^{20}}\)
=\(\frac{40!}{40!2^{20}}=\frac{1}{2^{20}}\)
So sánh \(U=\frac{1.3.5.7...39}{21.22.23...40}\) và V=\(\frac{1}{2^{20}-1}\)
Ai nhanh nhất mik tick cho
Ta có : \(U=\frac{1.3...39}{21.22...40}\)
=> \(U=\frac{1.3...39.\left(2.4...40\right)}{21.22...40.\left(2.4.6...40\right)}\)
=> \(U=\frac{1.2.3...39.40}{21.22...40.\left(1.2...20\right).2^{20}}\)
=> \(U=\frac{1}{2^{20}}\)
- Ta thấy : \(2^{20}>2^{20}-1\)
=> \(\frac{1}{2^{20}}< \frac{1}{2^{20}-1}\)
hay \(U< V\)
Vậy U < V .
So Sánh: U = \(\frac{1.3.5.7...39}{21.22.23...40}\)và V = \(\frac{1}{20^{20}-1}\)