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B = 9 . [ 4/1.3.5+4/3.5.7+4/5.7.9+...+4/25.27.29]
B = 9 . [ 1/3-1/783]
= 9 . [ 1/3-1/783]
= 9 . 260/783=260/87<261/87<3
a)\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
=\(\frac{9.4}{1.3.5}+\frac{9.4}{3.5.7}+\frac{9.4}{5.7.9}+...+\frac{9.4}{25.27.29}\)
=\(9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
=\(9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
=\(9.\left(\frac{1}{3}-\frac{1}{27.29}\right)=9.\left(\frac{1}{3}-\frac{1}{783}\right)=9.\left(\frac{261}{783}-\frac{1}{783}\right)=9.\frac{260}{783}\)
=\(\frac{260}{87}\)
b)
ta có: \(3=\frac{261}{87}>\frac{260}{87}\)
vậy A<3
Ta có :
\(B=\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+.............+\dfrac{36}{25.27.29}\)
\(B=9\left(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+...........+\dfrac{4}{25.27.29}\right)\)
\(B=9\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{5.7}+.............+\dfrac{1}{25.27}-\dfrac{1}{27.29}\right)\)
\(B=9\left(\dfrac{1}{1.3}-\dfrac{1}{27.29}\right)\)
\(B=9\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\)
\(B=9.\dfrac{1}{3}-9.\dfrac{1}{783}\)
\(B=3-\dfrac{9}{783}< 3\)
\(\Rightarrow B< 3\rightarrowđpcm\)
\(A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{25.27.29}\right)\)
\(A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)
\(A=9.\frac{260}{87}=\frac{260}{87}