tính hợp lí :
a) 2004 . 2004 -2003 . 2005
b) \(\frac{3}{4}.\frac{2}{5}-\frac{3}{10}.\frac{4}{3}\)
c) \(3\frac{1}{11}.\frac{27}{46}.1\frac{6}{7}.2\frac{4}{9}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(=\frac{1\frac{25}{108}.320\frac{1}{25}+46\frac{3}{4}}{4\frac{16}{21}:\left(-1\frac{20}{21}\right)}=\frac{330\frac{1}{25}}{-2\frac{18}{41}}=\)\(-135,3164\)
Ta có:
\(\frac{1\div2003+1\div2004-1\div2005}{5\div2003+5\div2004-5\div2005}\) - \(\frac{2\div2002+2\div2003-2\div2004}{3\div2002+3\div2003-3\div2004}\)
Đơn giản đi hết ta sẽ còn:
\(\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
2.
Ta có:
Số khoảng cách của các số trong dãy là 23 = 8
=> Tổng của dãy dưới sẽ gấp 8 lần tổng dãy trên.
=> 3025 . 8 = 24200
a)\(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{117}{12}-\frac{2}{12}=\frac{115}{12}\)
b)\(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)
c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)
a. \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{39}{4}-\frac{1}{6}=\frac{115}{12}\)
b. \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-1\frac{1}{2}\)
= \(\frac{13}{4}.1-\frac{3}{2}=\frac{13}{4}-\frac{3}{2}=\frac{7}{4}\)
c. \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2004}\right)\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2003}{2004}=\frac{1}{2004}\)
c.\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:-\frac{41}{21}}\)
\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}\)
\(\frac{100}{-\frac{100}{41}}=-41\)
a. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)
\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)
\(7:-\frac{1}{7}=-49\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Ta có:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)