(\(\frac{1}{1\cdot51}\)+\(\frac{1}{2\cdot52}\)+.......+\(\frac{1}{10\cdot60}\))x=(\(\frac{1}{1\cdot11}\)+\(\frac{1}{2\cdot12}\)+........+\(\frac{1}{40\cdot50}\))
Giải phương trình
\(\left(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+\cdot\cdot\cdot+\frac{1}{10\cdot60}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+\cdot\cdot\cdot+\frac{1}{50\cdot60}\)
Giải phương trình
\(\left(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+\cdot\cdot\cdot+\frac{1}{10\cdot60}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+\cdot\cdot\cdot+\frac{1}{50\cdot60}\)
giải phương trình
a,(\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{9\cdot10}\))(x-1)+\(\frac{1}{10}x\)=\(x-\frac{9}{10}\)
b,\(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+\frac{40}{39}\)
c,(x-10)+(x-19)+(x-18)+...+100+101=101
d,(\(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+...+\frac{1}{10\cdot60}\))x=\(\frac{1}{1\cdot11}+\frac{1}{1\cdot12}+\frac{1}{1\cdot13}\)
\(\frac{1}{10\cdot11}\)+\(\frac{1}{11\cdot12}\)+\(\frac{1}{12\cdot13}\)+ ......................+\(\frac{1}{49\cdot50}\)
Ta có: \(\frac{1}{10.11}+\frac{1}{11.12}+....+\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{10}-\frac{1}{50}\)
\(=\frac{2}{25}\)
Tìm x biết:
\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+...+\frac{1}{10\cdot110}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
Tìm số nguyên x biết:
\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
tìm x
\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+...+\frac{1}{10\cdot110}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
Tìm x
\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+...+\frac{1}{10\cdot110}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
Tính tổng: A=\(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\)
= \(\frac{1}{100}\left(\frac{1}{1}-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
=\(\frac{1}{100}\left(\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right)\)(1)
B = \(\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
=\(\frac{1}{10}\left(1+\frac{1}{2}+..+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-...-\frac{1}{110}\right)\)
=\(\frac{1}{10}\left(\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right)\) (2)
Từ (1) và (2) => x = B/A = 1/10 / 1/100 = 10