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Ta có: S = \(\dfrac{1}{3}+\dfrac{3}{3.7}+\dfrac{5}{3.7.11}+...+\dfrac{2n+1}{3.7.11...\left(4n+3\right)}\)
⇒ 2S = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+2}{3.7.11...\left(4n+3\right)}\)
⇒ 2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+3}{3.7.11...\left(4n+3\right)}\)
Đến đây nó sẽ rút gọn liên tục và sau nhiều lần rút gọn ta có:
2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+\dfrac{1}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{11}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{1}{3.7}\) = \(\dfrac{2}{3}+\dfrac{7}{3.7}=\dfrac{2}{3}+\dfrac{1}{3}=1\)
Suy ra 2S < 1 ⇒ S < \(\dfrac{1}{2}\)(đpcm)
8C=8/3.7.11+8/7.11.15+8/11.15.19+8/15.19.23+8/19.23.27+8/23.27.31
8C=1/3.7-1/7.11+1/7.11-1/11.15+1/11.15-1/15.19+1/15.19-1/19.23+1/19.23+-1/23.27-1/27.31
8C=1/3.7-1/27.31
8C=1/21-1/837
8C=272/5859
C=34/5859
4x-(7+5x)=15+(-29). b,x/11=x+4/33
4x-7-5x=-14. 33x=11(x+4)
4x-5x=14+7. 33x=11x+44
-x=21. 33x-11x=44
x=-21 22x=44
x=44:22=2
a: \(=\dfrac{21\cdot11}{22\cdot9}=\dfrac{1}{2}\cdot\dfrac{7}{3}=\dfrac{7}{6}\)
b: \(=\dfrac{49\cdot8}{10}=49\cdot\dfrac{4}{5}=\dfrac{196}{5}\)
c: \(=\dfrac{12\cdot\left(-4\right)}{32\cdot6}=\dfrac{-48}{192}=-\dfrac{1}{4}\)
Bài 1:
a: =-3/7
b: =-3/5
c: =1/13
d: =-1/2
Bài 2:
a: \(=\dfrac{21\cdot11}{22\cdot9}=\dfrac{1}{2}\cdot\dfrac{7}{3}=\dfrac{7}{6}\)
b: \(=\dfrac{49\cdot8}{10}=\dfrac{196}{5}\)
\(\dfrac{-63}{72}=\dfrac{-7}{8}\)
\(\dfrac{20}{-140}=\dfrac{-1}{7}\)
\(\dfrac{3.10}{5.24}=\dfrac{30}{120}=\dfrac{1}{4}\)
\(\dfrac{3.7.11}{22.9}=\dfrac{231}{198}=\dfrac{7}{6}\)
\(\dfrac{8.5-8.2}{16}=\dfrac{24}{16}=\dfrac{3}{2}\)
\(\dfrac{11.4-11}{2-13}=\dfrac{33}{-11}=-3\)