tính hợp lí :
a) 2004 . 2004 -2003 . 2005
b) \(\frac{3}{4}.\frac{2}{5}-\frac{3}{10}.\frac{4}{3}\)
c) \(3\frac{1}{11}.\frac{27}{46}.1\frac{6}{7}.2\frac{4}{9}\)
Tính:
a) \(A=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b)\(B=3-3^2+3^3-3^4+...-3^{2002}+3^{2003}-3^{2004}\)
a, \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(=\frac{1\frac{25}{108}.320\frac{1}{25}+46\frac{3}{4}}{4\frac{16}{21}:\left(-1\frac{20}{21}\right)}=\frac{330\frac{1}{25}}{-2\frac{18}{41}}=\)\(-135,3164\)
Bài 1 :
1 . Tính :
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
2 . Biết : 13 + 23 + ... + 103 = 3025
Tính : S = 23 + 43 + 63 + .... + 203
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
Tính bằng cách hợp lí ( Nếu có thể )
a) \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}\)
b) \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}\)
c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)\)
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. \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2004}\right)\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2003}{2004}=\frac{1}{2004}\)
Giải :
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) \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)
c) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)
1. Tính : P =\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{5}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
Tính : P = \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
Tìm x biết
a) (8-5x).(x+2)+4.(x-2).(x+1)+2.(x-2).(x+2)=0
b)\(\left(-\frac{2}{5}+x\right):\frac{7}{9}+\left(-\frac{3}{5}+\frac{5}{6}\right):\frac{7}{9}=0\)
c)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2004}\)
tính :
a) \(\frac{4}{9}:\frac{-1}{7}+6\frac{5}{9}:\frac{-1}{7}\)
b) \(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
c) \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
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. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)
\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)
\(7:-\frac{1}{7}=-49\)
b. \(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
\(\left(\frac{7}{3}+\frac{7}{2}\right):\left(-\frac{25}{6}+\frac{22}{7}\right)+\frac{15}{2}\)
\(\frac{35}{6}:-\frac{43}{42}+\frac{15}{2}\)
\(-\frac{245}{43}+\frac{15}{2}=\frac{155}{86}\)
Tính :
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Ta có:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
1/tính nhanh
a/1904.(2004-57)-2004.(1904-57)
b/\((\frac{1}{2}-1)\times(\frac{1}{3}-1)\times(\frac{1}{4}-1)\times...\times(\frac{1}{2004}-1)\)1)
c/\(\left(1+2^4+2^8\right):\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}+2^{11}\right)\)