a Tìm x , biết : 1\(\frac{3}{5}\) + [ \(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\)] x = \(\frac{16}{5}\) b Chứng minh rằng số tự nhiên A chia hết cho 2009 , với A = 1 . 2 .3 ... 2007 . 2008 ( 1 + \(\frac{1}{2}\) + ... + \(\frac{1}{2007}\)+ \(\frac{1}{2008}\)) Giảia...
Đọc tiếp
a Tìm x , biết : 1\(\frac{3}{5}\) + [ \(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\)] x = \(\frac{16}{5}\)
b Chứng minh rằng số tự nhiên A chia hết cho 2009 , với
A = 1 . 2 .3 ... 2007 . 2008 ( 1 + \(\frac{1}{2}\) + ... + \(\frac{1}{2007}\)+ \(\frac{1}{2008}\))
Giải
a 1\(\frac{3}{5}\)+ (\(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\)) x = \(\frac{16}{5}\)\(\Leftrightarrow\) \(\frac{8}{5}\)+ [\(\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\)x = \(\frac{16}{5}\)
\(\Leftrightarrow\)\(\frac{8}{5}\) + \(\frac{2}{5}\)x = \(\frac{16}{5}\)\(\Leftrightarrow\)\(\frac{2}{5}\)x = \(\frac{16}{5}\)\(-\)\(\frac{8}{5}\) \(\Leftrightarrow\) x = \(\frac{2}{5}\)x \(\Leftrightarrow\)= \(\frac{8}{5}\) : \(\frac{2}{5}\)\(\Leftrightarrow\)x=4
b 1 + \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ ... + \(\frac{1}{2007}\)+ \(\frac{1}{2008}\)
= (1 + \(\frac{1}{2008}\)) + (\(\frac{1}{2}\)+ \(\frac{1}{2007}\)) + ... + (\(\frac{1}{2004}\)+ \(\frac{1}{2005}\))
= (1 + \(\frac{1}{2008}\)) + (\(\frac{1}{2}\)+ \(\frac{1}{2007}\)) + ... + (\(\frac{1}{1004}\)+ \(\frac{1}{1005}\))
= \(\frac{2009}{1\times2008}\) + \(\frac{2009}{2\times2007}\) + ... + \(\frac{2009}{1004\times1009}\)
= 2009(\(\frac{1}{1\times2008}\) + \(\frac{1}{2\times2007}\)+ ... + \(\frac{1}{1004\times1005}\)
Do đó A = 1 . 2 .3 ... 2007 . 2008 . (1 + \(\frac{1}{2}\) + \(\frac{1}{3}\) + ... + \(\frac{1}{2007}\)+ \(\frac{1}{2008}\))
= 2009(1 . 2 . 3 ... 2007 . 2008 (\(\frac{1}{1.2008}\) + \(\frac{1}{2.2007}\)+ ... + \(\frac{1}{1004.1005}\) ) \(⋮\) 2009
Vì 1 . 2 . 3 ... 1007 . 2008 (\(\frac{1}{1.2008}\) + \(\frac{1}{2.2007}\) + ... + \(\frac{1}{2004.2005}\)) là một số tự nhiên
CÁC BẠN CÓ AI GIỐNG CÁCH LÀM CỦA MÌNH THÌ TRẢ LỜI NHÉ
\(\frac{223.2007+1034}{224.2007-973}\)
= \(\frac{223.2007+1034}{223.2007+2007-973}\)
= \(\frac{1034}{1034}\)= 1
\(\frac{223\times2007+1034}{224\times2007-973}\)
\(=\frac{223\times2007+1034}{\left(223+1\right)\times2007-973}\)
\(=\frac{223\times2007+1034}{223\times2007+2007-973}\)
\(=\frac{223\times2007+1034}{223\times3007+1034}\)
\(=1\)