Cho B = \(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)
So sánh B với \(\frac{1}{2}\)
So sánh tổng S=\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)với \(\frac{1}{2}\)
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
So sánh: \(S=\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)với \(\frac{1}{2}\)
SO SÁNH :S=\(\frac{1}{5}\)+\(\frac{1}{9}\)+\(\frac{1}{10}\)+\(\frac{1}{41}\)+\(\frac{1}{42}\)với \(\frac{1}{2}\)
\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}
\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)
\(< \frac{1}{5}+\frac{1}{8}+\frac{1}{8}+\frac{1}{40}+\frac{1}{40}\)
\(=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
1. so sánh
A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+với1\)
B=\(1-\left(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}\right)với\frac{1}{2}\)
C=\(1-\left(\frac{1}{5}+\frac{1}{11}+\frac{1}{10}+\frac{1}{9}+\frac{1}{59}+\frac{1}{58}+\frac{1}{57}\right)với\frac{1}{2}\)
Tính tỏng\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(A=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\) so sánh a với \(\frac{7}{42}\)
\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+..+\frac{2}{37\cdot38\cdot39}\)
\(2B=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+..+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2B=\frac{1}{2}-\frac{1}{1482}\)
\(B=\frac{185}{741}\)
\(A>\frac{1}{80}+\frac{1}{80}+..+\frac{1}{80}\)
\(A>\frac{1}{80}\cdot40>\frac{7}{42}\)
\(A>\frac{7}{42}\)
So sánh tổng S=\(\frac{1}{5}\)+\(\frac{1}{9}\)+\(\frac{1}{10}\)+\(\frac{1}{41}\)+\(\frac{1}{42}\) với \(\frac{1}{2}\)
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
So sánh :
a)\(\frac{3}{124},\frac{1}{41},\frac{5}{207},\frac{2}{83}\)
b)\(\frac{-2525}{2929}và\frac{-217}{245}\)
c)\(A=\frac{3^{10}+1}{3^9+1}vàB=\frac{3^9+1}{3^8+1}\)
d)\(\frac{27}{82}và\frac{26}{75}\)
Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
B = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
a) So sánh A và B
b) Chứng minh A = \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
So sánh \(A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\) và \(B=\frac{7}{12}\) ta được kết quả là:...
Giúp mình với mình cần gấp!
Ta có:
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80
1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60
và 1/61> 1/62> ... >1/79> 1/80
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80
Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12
=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12
\(A>B\)