\(\dfrac{1}{1}-\dfrac{1}{2000}=\dfrac{2000}{2000}-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
\(\dfrac{1}{1}-\dfrac{1}{2000}=\dfrac{2000}{2000}-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
a, \(\dfrac{1}{1}-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
Chúc bạn học tốt!!!
\(\dfrac{1}{1}-\dfrac{1}{2000}=\dfrac{2000}{2000}-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
\(\dfrac{1}{1}-\dfrac{1}{2000}=\dfrac{2000}{2000}-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
a, \(\dfrac{1}{1}-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
Chúc bạn học tốt!!!
BT3: Tìm x, biết
20) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=\dfrac{1998}{2000}\)
Câu hỏi không hay nhưng có thưởng vài điểm cho các bạn nè😁😁😁
Đề bài: Tìm \(x\), biết:
\(1,\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{23}{45}\)
\(2,\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{1998}{2000}\)
Bài 1 : Tính bằng cách hợp lí :
a ) A = \((\) \(\dfrac{\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}}{\dfrac{2}{13}-\dfrac{2}{17}-\dfrac{2}{23}}\) \()\) . \((\) \(\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}-0,875+0,7}\) \()\) + \(\dfrac{6}{7}\)
b ) B = 2000 : \([\) \((\) \(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}\) \()\) . \((\) \(\dfrac{-1\dfrac{1}{6}+0,875-0,7}{\dfrac{1}{3}-0,25+\dfrac{1}{5}}\) \()\) \(]\)
c ) C = 10101 . \((\) \(\dfrac{5}{111111}+\dfrac{2}{222222}-\dfrac{4}{21.143.37}\) \()\)
Bài 2 : Tìm x :
a ) \(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+......+\dfrac{1}{125.155}-x=\dfrac{1}{310}\)
b ) \(1\dfrac{3}{5}+(\dfrac{\dfrac{2}{7}\dfrac{2}{17}\dfrac{2}{37}}{\dfrac{5}{7}\dfrac{5}{17}\dfrac{5}{37}}).\) x = \(\dfrac{16}{5}\)
\(\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{5}-1\right).\left(\dfrac{1}{6}-1\right)...\left(\dfrac{1}{2000}-1\right).\left(\dfrac{1}{2001}-1\right)\)
Mn giúp mk đi mk cần gấp lắm làm ơn !
1) Rút gọn
A =\(\dfrac{\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+.......+\dfrac{18}{2}+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.......+\dfrac{1}{19}+\dfrac{1}{20}}\)
2) Tìm x
a/ \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{x.\left(x+1\right)}=\dfrac{2016}{2017}\)
Cho A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{2}{98}+\dfrac{1}{99}}\)
Tính A
BT1: CMR:
a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)
b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)
c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\)
d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
BT2: Tính tổng
a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\)
BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)
CMR: 1 < S < 2
Tính hợp lí A = \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}\)
1 Tính:
a)\(\dfrac{\dfrac{7}{10}+\dfrac{3}{5}}{\dfrac{7}{10}+\dfrac{1}{2}}\) b)\(\dfrac{6-\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}}{6+\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}}\)