So sánh tổng 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 \(\frac{1}{2}\)
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 :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}
\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)
\(< \frac{1}{5}+\frac{1}{8}+\frac{1}{8}+\frac{1}{40}+\frac{1}{40}\)
\(=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
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
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 tổng dãy số: \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+..+\frac{1}{80}\)và so sánh với \(\frac{7}{12}\)
So sanh S=\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{40}+\frac{1}{42}voi\frac{1}{2}\)
minh can cach lam
ai nhanh minh tick
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
1/5+1/9+1/10+1/40+1/42=1159/2520
1159/2520=0.4599.....
1/2=0.5
Mà:0.5>0.4599
Nên:1/5+1/9+1/10+1/40+1/42>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 :
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)\(\frac{27}{82}và\frac{26}{75}\)