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\(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+............+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{\left(2.4\right)^2}+\frac{6^2-4^2}{\left(4.6\right)^2}+..........+\frac{98^2-96^2}{\left(96.98\right)^2}+\frac{100^2-98^2}{\left(98.100\right)^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-...............-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{10000}\)
\(B=\frac{2500}{10000}-\frac{1}{10000}\)
\(B=\frac{2499}{10000}\)
Vậy B = \(\frac{2499}{10000}\)
Ta có :
\(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(=\frac{12}{4.16}+\frac{20}{16.36}+...+\frac{388}{9216.9604}+\frac{396}{9604.10000}\)
\(=\frac{1}{4}-\frac{1}{16}+\frac{1}{16}-\frac{1}{36}+...+\frac{1}{9604}-\frac{1}{10000}\)
\(=\frac{1}{4}-\frac{1}{10000}< \frac{1}{4}\)
\(\Leftrightarrow B< \frac{1}{4}\)
B=\(\frac{12}{4.16}\)+\(\frac{20}{16.36}\)+...+\(\frac{396}{9604.10000}\)
Ta có:\(\frac{12}{4.16}\)=\(\frac{1}{4}\)-\(\frac{1}{16}\)
\(\frac{20}{16.36}\)=\(\frac{1}{16}\)-\(\frac{1}{36}\)
...
Khi đó:B=\(\frac{1}{4}\)-\(\frac{1}{16}\)+\(\frac{1}{16}\)-\(\frac{1}{36}\)+...+\(\frac{1}{9604}\)-\(\frac{1}{10000}\)=\(\frac{1}{4}\)-\(\frac{1}{10000}\)<\(\frac{1}{4}\)
Vậy: B<\(\frac{1}{4}\)
Ta có:
B=\(\frac{4^2-2^2}{2^2\times4^2}+\frac{6^2-4^2}{4^2\times6^2}+...+\frac{98^2-96^2}{96^2\times98^2}+\frac{100^2-98^2}{98^2\times100^2}\)
=\(\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
= \(\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
B = \(\dfrac{12}{\left(2.4\right)^2}+\dfrac{20}{\left(4.6\right)^2}+...+\dfrac{388}{\left(96.98\right)^2}+\dfrac{396}{\left(98.100\right)^2}\)
= \(\dfrac{4^2-2^2}{2^{2^{ }}.4^{2^{ }}}+\dfrac{6^{2^{ }}-4^2}{4^2.6^2}+...+\dfrac{98^2-96^2}{96^2.98^2}+\dfrac{100^2-98^2}{98^2.100^2}\)
=\(\dfrac{1}{2^{2^{ }}}-\dfrac{1}{4^{2^{ }}}+\dfrac{1}{4^2}-\dfrac{1}{6^2}+\dfrac{1}{6^2}+....-\dfrac{1}{98^2}+\dfrac{1}{98^2}-\dfrac{1}{100^2}\)
= \(\dfrac{1}{2^2}-\dfrac{1}{100^2}=\dfrac{1}{4}-\dfrac{1}{100^2}< \dfrac{1}{4}\)
Vậy B < \(\dfrac{1}{4}\)
B = 12(2.4)2+20(4.6)2+...+388(96.98)2+396(98.100)212(2.4)2+20(4.6)2+...+388(96.98)2+396(98.100)2
= 42−2222.42+62−4242.62+...+982−962962.982+1002−982982.100242−2222.42+62−4242.62+...+982−962962.982+1002−982982.1002
=122−142+142−162+162+....−1982+1982−11002122−142+142−162+162+....−1982+1982−11002
= 122−11002=14−11002<14122−11002=14−11002<14
Vậy B < 14
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{100}{2}=50\)
\(\frac{-17}{2.4}-\frac{17}{4.6}-\frac{17}{6.8}-...-\frac{17}{100.102}\)
\(=-\frac{17}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)
\(=-\frac{17}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(=-\frac{17}{2}\left(\frac{1}{2}-\frac{1}{102}\right)\)
\(=-\frac{17}{2}.\frac{25}{51}=-\frac{25}{6}\)
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
a,\(\left(3-2\frac{1}{3}+\frac{1}{4}\right):\left(4-5\frac{1}{6}+2\frac{1}{4}\right)\) =\(\left(3-\frac{7}{3}+\frac{1}{4}\right):\left(4-\frac{31}{6}+\frac{9}{4}\right)\) =\(\left(3-\frac{31}{12}\right):\left(4-\frac{1}{3}\right)\) =\(\frac{5}{12}:\frac{11}{3}\) =\(\frac{5}{44}\) b, F=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.......+\frac{4}{2008.2010}\) =\(2.\left(1-\frac{2}{2010}\right)\) =\(2.\frac{1004}{1005}\) =\(\frac{2008}{1005}\)
\(A=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+\frac{28}{\left(6.8\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(100.98\right)^2}\)
\(=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+\frac{8^2-6^2}{6^2.8^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+\frac{1}{6^2}-\frac{1}{8^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(=\frac{1}{2^2}-\frac{1}{100^2}< \frac{1}{4}\)
A = \(\frac{12}{\left(2.4\right)^2}\) + \(\frac{20}{\left(4.6\right)^2}\) + ..........+ \(\frac{388}{\left(96.98\right)^2}\) + \(\frac{396}{\left(98.100\right)^2}\)
= \(\frac{16-4}{\left(2.4\right)^2}\)+ \(\frac{36-16}{\left(4.6\right)^2}\)+...........+ \(\frac{9604-9216}{\left(96.98\right)^2}\) + \(\frac{10000-9604}{\left(98.100\right)^2}\)
= \(\frac{1}{2^2}\) - \(\frac{1}{4^2}\)+ \(\frac{1}{4^2}\)- \(\frac{1}{6^2}\) + ............+ \(\frac{1}{96^2}\) - \(\frac{1}{98^2}\) + \(\frac{1}{98^2}\) - \(\frac{1}{100^2}\)
= \(\frac{1}{2^2}\) - \(\frac{1}{100^2}\)
= \(\frac{1}{4}\) - \(\frac{1}{100^2}\) < \(\frac{1}{4}\)
=) A < \(\frac{1}{4}\)
Chúc bạn học tốt nhé !