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B=\(\frac{12}{2^2.4^2}+\frac{20}{4^2.6^2}+......+\frac{388}{96^2.98^2}+\frac{396}{98^2.100^2}\)
=\(\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
=\(\frac{1}{2^2}-\frac{1}{100^2}\)
=\(\frac{2599}{10000}< \frac{2500}{10000}=\frac{1}{4}\)
=> B<\(\frac{1}{4}\)
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<\(\frac{1}{4}\)
\(\frac{3}{2\cdot4}+\frac{3}{4\cdot6}+\frac{3}{6\cdot8}+...+\frac{3}{96\cdot98}\)
\(=\frac{3}{2}\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{96\cdot98}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{1}{2}-\frac{1}{98}\right)=\frac{3}{2}\cdot\left(\frac{48}{98}-\frac{1}{98}\right)\)
\(=\frac{3}{2}\cdot\frac{47}{98}=\frac{141}{196}\)
Ta thấy \(10^{50}>10^{50}-3\)
\(\Rightarrow B=\frac{10^{50}}{10^{50}-3}>\frac{10^{50}+2}{10^{50}-3+2}=\frac{10^{50}+2}{10^{50}-1}=A\)
Vậy \(A< B\)
Tk mình đi mọi người mình bị âm nè!
Ai tk mình mình tk lại cho
B=\(\frac{12}{\left(2.4\right)^2}\)+\(\frac{20}{\left(4.6\right)^2}\)+.....+\(\frac{396}{\left(98.100\right)^2}\)
B=\(\frac{4^2-2^2}{2^2.4^2}\)+ \(\frac{6^2-4^2}{4^2.6^2}\)+....+\(\frac{100^2-98^2}{\left(98^2.100^2\right)}\)
B=\(\frac{1}{2^2}\)-\(\frac{1}{4^2}\)+\(\frac{1}{4^2}\)-\(\frac{1}{6^2}\)+....+\(\frac{1}{98^2}\)-\(\frac{1}{100^2}\)
B=\(\frac{1}{2^2}\)-\(\frac{1}{100^2}\)< \(\frac{1}{2^2}\)=\(\frac{1}{4}\)
Vậy B<\(\frac{1}{4}\)
B<\(\frac{1}{4}\)