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b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
Làm tiếp:
\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)
\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)
Bài 2:
Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)
\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)
\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)
Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
Bài 1:Tính
a, Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)
Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)
\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)
\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)
\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)
Áp dụng vào bài toán ta có đáp số là:1
b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)
c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)
d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)
e,Xét mẫu số ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)
\(M=\frac{3}{1^22^2}+\frac{5}{2^23^2}+\frac{7}{3^24^2}+...+\frac{4019}{2009^22010^2}\)
\(M=\frac{2^2-1^2}{1^22^2}+\frac{3^2-2^2}{2^23^2}+\frac{4^2-3^2}{3^24^2}+...+\frac{2010^2-2009^2}{2009^22010^2}\)
\(M=\frac{2^2}{1^22^2}-\frac{1^2}{1^22^2}+\frac{3^2}{2^23^2}-\frac{2^2}{2^23^2}+\frac{4^2}{3^24^2}-\frac{3^2}{3^24^2}+...+\frac{2010^2}{2009^22010^2}-\frac{2009^2}{2009^22010^2}\)
\(M=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}\)
\(M=1-\frac{1}{2010^2}< 1\)
Vậy \(M< 1\)
Chúc bạn học tốt ~
1.\(\left(-\frac{6}{5}+\frac{6}{16}-\frac{6}{23}\right):\left(\frac{9}{5}-\frac{9}{16}+\frac{9}{23}\right)\)
\(=6\left(-\frac{1}{5}+\frac{1}{16}-\frac{1}{23}\right):\left(-9\right)\left(\frac{-1}{5}+\frac{1}{16}-\frac{1}{23}\right)\)
\(=6:\left(-9\right)=-\frac{2}{3}\)
2. \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{0.5-\frac{1}{3}+\frac{1}{4}}{-\frac{3}{2}+1-\frac{3}{4}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{-3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}-\frac{1}{3}\)
\(=\frac{9}{13}-\frac{5}{15}=\frac{4}{15}\)
Ta có :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\)\(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\)\(\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+\frac{4^2}{3^2.4^2}-\frac{3^2}{3^2.4^2}+...+\frac{10^2}{9^2.10^2}-\frac{9^2}{9^2.10^2}\)
\(=\)\(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=\)\(1-\frac{1}{10^2}\)
\(=\)\(\frac{100-1}{100}\)
\(=\)\(\frac{99}{100}\)
Chúc bạn học tốt ~
=3/1.4+5/4.9+7/9.16+......+19/81.100
=(1/1-1/4)+(1/4-1/9)+........+(1/81-1/100)
=1-1/100
=99/100<1(đpcm)
a)Xét vế trái , ta có :
Gọi tổng các số hạng ở vế trái là A
=> A= \(\frac{1}{3}\)+\(\frac{1}{3^2}\)+ ... +\(\frac{1}{3^{99}}\)
=>3A = 1 + \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ ... + \(\frac{1}{3^{98}}\)
=> 3A - A = 1 + \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ ... + \(\frac{1}{3^{98}}\)- ( \(\frac{1}{3}\)+\(\frac{1}{3^2}\)+ ... +\(\frac{1}{3^{99}}\))
=> 2A = 1 - \(\frac{1}{3^{99}}\)
=> A = \(\frac{1}{2}\)- \(\frac{1}{3^{99}.2}\) < \(\frac{1}{2}\)
b)\(\frac{3}{1^2.2^2}\)+ \(\frac{5}{2^2.3^2}\)+ ... + \(\frac{19}{9^2.10^2}\)
= \(\frac{3}{1.4}\)+ \(\frac{5}{4.9}\)+ .... + \(\frac{19}{81.100}\)
= 1 - \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{9}\)+ ... + \(\frac{1}{81}\)- \(\frac{1}{100}\)
= 1 - \(\frac{1}{100}\) <1
a,
\(\sum\limits^{99}_{x=1}\left(\frac{1}{3^x}\right)=\frac{1}{2}\)
bài a nó có ............
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+.....+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+......+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+.....+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=\frac{1}{1^2}-\frac{1}{10^2}=1-\frac{1}{10^2}<1\left(đpcm\right)\)