So sánh tổng S=\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)với \(\frac{1}{2}\)
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\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}
\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)
\(< \frac{1}{5}+\frac{1}{8}+\frac{1}{8}+\frac{1}{40}+\frac{1}{40}\)
\(=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
S = \(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}=\frac{5932}{12915}=0.459310878\approx0.45\)
\(\frac{1}{2}=0.5\)
Vì 0.45 < 0.5
\(\Leftrightarrow\) S < 1/2
Nhận xét: \(\frac{1}{5}< \frac{1}{42};\frac{1}{9}< \frac{1}{42};\frac{1}{10}< \frac{1}{42};\frac{1}{40}< \frac{1}{42}\)
\(\Rightarrow S< \frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}\)
\(\Rightarrow S< \frac{5}{42}< \frac{21}{42}=\frac{1}{2}\)
Vậy S < 1/2
\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+..+\frac{2}{37\cdot38\cdot39}\)
\(2B=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+..+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2B=\frac{1}{2}-\frac{1}{1482}\)
\(B=\frac{185}{741}\)
\(A>\frac{1}{80}+\frac{1}{80}+..+\frac{1}{80}\)
\(A>\frac{1}{80}\cdot40>\frac{7}{42}\)
\(A>\frac{7}{42}\)
bé hơn nha bạn
Ta có: 1/9 + 1/10 < 1/8+1/8 = 1/4
1/41+1/42< 1/40+1/40=1/20
=> 1/5+1/9+1/10+1/41+1/42<1/5+1/4+1/20=1/2
Vậy 1/5+1/9+1/10+1/41!+1/42<1/2