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\(b)\) Đặt \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\) ta có :
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}-0=\frac{1}{2}\)
\(\Rightarrow\)\(A< \frac{1}{2}\) ( đpcm )
Vậy \(A< \frac{1}{2}\)
Chúc bạn học tốt ~
\(a)\frac{9.25-63}{3.30+153}\)
\(=\frac{9.25-9.7}{3.30+3.51}\)
\(=\frac{9.\left(25-7\right)}{3.\left(30+51\right)}\)
\(=\frac{9.18}{3.81}\)
\(=\frac{1.6}{1.9}\)
\(=\frac{6}{9}\)
\(=\frac{2}{3}\)
b ) \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(Đpcm\right)\)
Chúc bạn học tốt !!!
\(\frac{2.4}{6.18}=\frac{1.2}{3.9}=\frac{2}{27}\)
\(\frac{3.5.7}{6.9.14}=\frac{1.5.7}{2.9.2}=\frac{35}{36}\)
\(\frac{4.7-4.5}{64}=\frac{4\left(7-5\right)}{64}=\frac{8}{64}=\frac{1}{8}\)
a. \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2013}}\)
\(\Rightarrow3A-A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{2014}}\right):2=\frac{1}{2}-\frac{1}{3^{2014}.2}=\frac{3^{2014}-1}{3^{2014}.2}\)
b.\(B=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}\)
\(\Rightarrow2B=1+\frac{1}{2^2}+....+\frac{1}{2^{2013}}\)
\(\Rightarrow2B-B=1-\frac{1}{2^{2014}}\)
\(\Rightarrow B=1-\frac{1}{2^{2014}}\)
Bài 1 : \(\frac{x-12}{4}=\frac{1}{2}\)
\(\Rightarrow2\cdot(x-12)=1\cdot4\)
\(\Rightarrow2x-24=4\)
\(\Rightarrow2x=28\)
\(\Rightarrow x=14\)
Vậy x = 14
Bài 2 : Rút gọn phân số
\(a,\frac{-315}{540}=\frac{-7}{12}\)
\(b,\frac{25\cdot13}{26\cdot35}=\frac{5\cdot1}{2\cdot7}=\frac{5}{14}\)
\(c,\frac{6\cdot9-2\cdot17}{63\cdot3-119}=\frac{54-34}{70}=\frac{20}{70}=\frac{2}{7}\)
\(d,\frac{1989\cdot1990+3978}{1992\cdot1991-3984}=1\)
Bài 3 tự so sánh nhé :v
d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)
\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)
\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)
\(=-\frac{2018}{2018}\)
\(=-1\)
Ta có : \(A=\frac{1}{1.101}+\frac{1}{2.202}+\frac{1}{3.103}+...+\frac{1}{10.110}\)
=\(\frac{1}{100}.\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right)\)
= \(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right)\)
= \(\frac{1}{100}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)
Lại có : B = \(\frac{1}{10}.\left(\frac{10}{1.11}+\frac{10}{2.12}+\frac{10}{3.13}+...+\frac{10}{100.110}\right)\)
= \(\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{110}\right)\)
= \(\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)
Khi đó \(A:B=\frac{A}{B}=\frac{\frac{1}{100}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}{\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}=\frac{1}{10}\)
\(A=\frac{15.39}{9.25}=\frac{3.5.3.13}{3.3.5.5}=\frac{13}{5}\)
\(B=\frac{21-21.5}{2^2}=\frac{21.\left(1-5\right)}{4}=\frac{21.\left(-4\right)}{4}=-21\)
\(C=\frac{1.2.3....20}{2.4...40}=\frac{1.2.3...20}{2\left(1.2.3...20\right)}=\frac{1}{2}\)
Ta có :
\(A=\frac{15.39}{9.25}=\frac{3.5.39}{3^2.5^2}=\frac{39}{3.5}=\frac{39}{15}\)
\(B=\frac{21-21.5}{2^2}=\frac{21.\left(1-5\right)}{4}=\frac{21.\left(-4\right)}{4}=\frac{21.\left(-1\right)}{1}=-21\)
\(C=\frac{1.2.3...20}{2.4.6...40}=\frac{\left(1.3.5...19\right).\left(2.4.6...20\right)}{\left(2.4.6...20\right).\left(22.24.26...40\right)}=\frac{1.3.5.19}{22.24.26...40}\)