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\(\frac{1}{10}.\frac{1}{11}+\frac{1}{11}.\frac{1}{12}+\frac{1}{12}.\frac{1}{13}+....+\frac{1}{20}.\frac{1}{21}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+....+\frac{1}{20}-\frac{1}{21}\)
\(=\frac{1}{10}-\frac{1}{21}\)
\(=\frac{11}{210}\)
Đặt A=1.2+2.3+3.4+............+99.100
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+.............+99.100.(101-98)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+................+99.100.101-98.99.100
3A=99.100.101
A=99.100.101:3
A=333300
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ < \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{97}{98}.\dfrac{98}{99}< \dfrac{1}{99}\\ < \dfrac{1}{10}.\\\\ =>A< \dfrac{1}{10}\)
a) \(\frac{2}{3}x\times\frac{1}{2}=\frac{1}{10}\Rightarrow\frac{2}{3}x=\frac{1}{5}\Rightarrow x=\frac{3}{10}\)
(1 + 1/2) . (1 + 1/3) . (1 + 1/4) ... (1 + 1/100)
= 3/2 . 4/3 . 5/4 ... 101/100
= 101/2
3/2 . 4/3 . 5/4 . .... . 101/100
101/2
pạn rút gọn sẽ thấy đúng thôi
\(\frac{10}{3}.x+\frac{67}{4}=\frac{53}{4}\)
\(\frac{10}{3}.x=\frac{53}{4}-\frac{67}{4}\)
\(\frac{10}{3}.x=-\frac{7}{2}\)
\(x=-\frac{7}{2}:\frac{10}{3}\)
\(x=-\frac{21}{20}\)
a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{100\cdot101}\)
=1-1/2+1/2-1/3+...+1/100-1/101
=1-1/101=100/101
b: \(A=1+\dfrac{1}{2}+1+\dfrac{1}{6}+1+\dfrac{1}{12}+...+1+\dfrac{1}{10100}\)
\(=100+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)
\(=101-\dfrac{1}{101}< 101\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{100}{609}\\ \)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{1}{3}-\frac{1}{2x+3}\)\(=\frac{2x}{3\left(2x+3\right)}\)
\(A=\frac{x}{3\left(2x+3\right)}=\frac{100}{609}=\frac{100}{3.203}=\frac{100}{3\left(2.100+3\right)}\)\(\Rightarrow x=100\)
\(A=\left(\frac{1}{10}-1\right).\left(\frac{1}{11}-1\right)....\left(\frac{1}{100}-1\right)\))
\(A=\frac{-9}{10}.\frac{-10}{11}...\frac{-99}{100}\) ( lẻ thừa số nguyên âm)
\(A=\frac{-9}{100}\)
\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)...\left(\frac{1}{100}-1\right)\)
\(\Rightarrow A=-\frac{9}{10}.\frac{-10}{11}...\frac{-99}{100}\)
\(\Rightarrow-A=\frac{9}{10}.\frac{10}{11}...\frac{99}{100}\)
\(\Rightarrow-A=\frac{9.10.11...99}{10.11...99.100}\)
\(\Rightarrow-A=\frac{9}{100}\)
\(\Rightarrow A=-\frac{9}{100}\)