So sanh cac phan so sau:
A=\(\frac{3^{10}+1}{3^9+1}\)va B=\(\frac{3^9+1}{3^8+1}\)
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H
2 tháng 1 2018
\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
NN
1
DL
3 tháng 5 2019
\(A=\frac{3^{10}+1}{3^9+1}=\frac{3^{10}+3-2}{3^9+1}=\frac{3\left(3^9+1\right)-2}{3^9+1}=3-\frac{2}{3^9+1}\)
\(B=\frac{3^9+1}{3^8+1}=\frac{3^9+3-2}{3^8+1}=\frac{3\left(3^8+1\right)-2}{3^8+1}=3-\frac{2}{3^8+1}\)
Có \(3^9+1>3^8+1\)
\(\Rightarrow\frac{2}{3^9+1}< \frac{2}{3^8+1}\)
\(\Rightarrow3-\frac{2}{3^9+1}>3-\frac{2}{3^8+1}\)
\(\Rightarrow A>B\)
Q
23 tháng 6 2017
a) Ta có:
+) \(\dfrac{1}{2}=\dfrac{3}{6}\)
+) \(\dfrac{1}{3}=\dfrac{2}{6}\)
+) \(\dfrac{2}{3}=\dfrac{4}{6}\)
=> \(\dfrac{2}{6}< \dfrac{3}{6}< \dfrac{4}{6}\)
hay \(\dfrac{1}{3}< \dfrac{1}{2}< \dfrac{2}{3}\)
b) Ta có:
+) \(\dfrac{4}{9}=\dfrac{56}{126}\)
+) \(-\dfrac{1}{2}=-\dfrac{63}{126}\)
+) \(\dfrac{3}{7}=\dfrac{54}{126}\)
=> \(-\dfrac{63}{126}< \dfrac{54}{126}< \dfrac{56}{126}\)
hay \(-\dfrac{1}{2}< \dfrac{3}{7}< \dfrac{4}{9}\)
c) Ta có:
+) \(\dfrac{27}{82}=\dfrac{2025}{6150}\)
+) \(\dfrac{26}{75}=\dfrac{2132}{6150}\)
=> \(\dfrac{2025}{6150}< \dfrac{2132}{6150}\)
hay \(\dfrac{27}{82}< \dfrac{26}{75}\)
d) Ta có:
+) \(-\dfrac{49}{78}=-\dfrac{4655}{7410}\)
+) \(-\dfrac{64}{95}=-\dfrac{4992}{7410}\)
=> \(-\dfrac{4665}{7410}>-\dfrac{4992}{7410}\)
hay \(-\dfrac{49}{78}>-\dfrac{64}{95}\)
LP
0
bn vào /h7.net/hoi-dap/toan-6/so-sanh-a-3-10-1-3-9-1-va-b-3-9-1-3-8-1--faq205231.html
Trả lời:
A = \(\frac{3^{10}+1}{3^9+1}=\frac{3.3^9+1}{3.3^8+1}=\frac{3^9+1}{3^8+1}\)= B
_Học tốt bạn nha_