so sánh:
M=10^30+1/10^31+1 và N=10^31+1/10^32+1
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Cách khác
\(E=\frac{10^{30}+2}{10^{31}+2}\Rightarrow10E=\frac{10^{31}+20}{10^{31}+2}=\frac{10^{31}+2+18}{10^{31}+2}=1+\frac{18}{10^{31}+2}\)
\(F=\frac{10^{31}+2}{10^{32}+2}\Rightarrow10F=\frac{10^{32}+20}{10^{32}+2}=\frac{10^{32}+2+18}{10^{32}+2}=1+\frac{18}{10^{32}+2}\)
Vì \(\frac{18}{10^{31}+2}>\frac{18}{10^{32}+2}\Rightarrow1+\frac{18}{10^{31}+2}>1+\frac{18}{10^{32}+2}\Rightarrow E>F\)
Ta co:
B=\(\frac{10^{30}+1}{10^{31}+1}\)<\(\frac{10^{30}+1+99}{10^{31}+1+99}\)=\(\frac{10^{30}+100}{10^{31}+100}\)=\(\frac{10^{10}\cdot\left(10^{20}+1\right)}{10^{10}\cdot\left(10^{21}+1\right)}\)=\(\frac{10^{20}+1}{10^{21}+1}\)=A
Vay A<B
số số hạng là :
( 1000000 - 1 ) : 1 + 1 = 1000000
tổng là :
( 1000000 + 1 ) x 1000000 : 2 = 500000500000
đáp số : 500000500000
Ta có : \(\frac{1}{9}>\frac{1}{16}\)
\(\frac{1}{10}>\frac{1}{16}\)
\(\frac{1}{11}>\frac{1}{16}\)
............
\(\frac{1}{16}=\frac{1}{16}\)
\(\Rightarrow\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{16}>\frac{1}{16}\times8=\frac{1}{2}\)
\(\frac{1}{17}>\frac{1}{32}\)
\(\frac{1}{18}>\frac{1}{32}\)
\(\frac{1}{19}>\frac{1}{32}\)
..........
\(\frac{1}{32}=\frac{1}{32}\)
\(\Rightarrow\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+...+\frac{1}{32}>\frac{1}{32}\times8=\frac{1}{4}\)
\(\Rightarrow\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{32}>\frac{1}{3}+\frac{1}{4}\)
\(M=\frac{10^{30}+1}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10\cdot(10^{30}+1)}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+10}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+1+9}{10^{31}+1}\)
\(\Rightarrow10M=1+\frac{9}{10^{31}+1}\)
\(N=\frac{10^{31}+1}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10\cdot(10^{31}+1)}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+10}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+1+9}{10^{32}+1}\)
\(\Rightarrow10N=1+\frac{9}{10^{32}+1}\)
Mà\(1+\frac{9}{10^{31}+1}>1+\frac{9}{10^{32}+1}\)
Nên \(10M>10N\)
Hay \(M>N\)