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Ta có \(\frac{1}{9S}=\frac{9^{2017}+\frac{1}{9}}{9^{2017}+1}\)= \(\frac{9^{2017}+1-\frac{8}{9}}{9^{2017}+1}=1-\frac{\frac{8}{9}}{9^{2017}+1}\)
\(\frac{1}{9M}=\frac{9^{2016}+\frac{1}{9}}{9^{2016}+1}\)= \(\frac{9^{2016}+1-\frac{8}{9}}{9^{2016}+1}=1-\frac{\frac{8}{9}}{9^{2016}+1}\)
Vì \(9^{2016}+1< 9^{2017}+1\)=> \(\frac{\frac{8}{9}}{9^{2016}+1}>\frac{\frac{8}{9}}{9^{2017}+1}\)
=> \(1-\frac{\frac{8}{9}}{9^{2016}+1}< 1-\frac{\frac{8}{9}}{9^{2017}+1}\)=> \(\frac{1}{9}S< \frac{1}{9}M\Rightarrow S< M\)
a)\(4^{72}=\left(4^3\right)^{24}=64^{24}\)
\(8^{48}=\left(8^2\right)^{24}=64^{24}\)
\(\Rightarrow4^{72}=8^{48}\)
a) \(4^{72}=\left(2^2\right)^{72}=2^{144}\)
\(8^{48}=\left(2^3\right)^{48}=2^{144}\)
mà \(2^{144}=2^{144}\)=> \(4^{72}=8^{48}\)
b) \(2^{252}=\left(2^2\right)^{126}=4^{126}\)
mà \(4^{126}< 5^{127}\)=> \(5^{127}>2^{252}\)
bài 1 :
\(\frac{2}{3}\)+\(\frac{1}{3}\)=\(\frac{3}{3}\)=1
\(\frac{3}{4}\)+\(\frac{2}{4}\)+\(\frac{1}{4}\)=\(\frac{4}{4}\)=1
\(\frac{4}{5}\)+\(\frac{3}{5}\)+\(\frac{2}{5}\)+\(\frac{1}{5}\)=\(\frac{10}{5}\)= 2
chúc bạn học tốt !!!
Ta có : \(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}=3\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)=3\left(\frac{1}{20}-\frac{1}{80}\right)=3.\frac{3}{80}=\frac{9}{80}< 1\)
\(\frac{2}{3}+\frac{1}{3}=\frac{6+3}{3}=\frac{9}{3}=3\)
\(\frac{3}{4}+\frac{2}{4}+\frac{1}{4}=\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{1}{2}=1+\frac{1}{2}=1\frac{1}{2}=\frac{3}{2}\)
\(\frac{4}{5}+\frac{3}{5}+\frac{2}{5}+\frac{1}{5}=\left(\frac{4}{5}+\frac{1}{5}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)=2+2=4\)
\(\frac{5}{6}+\frac{4}{6}+\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\left(\frac{5}{6}+\frac{1}{6}\right)+\left(\frac{4}{6}+\frac{2}{6}\right)+\frac{1}{2}=1+1\)\(+\frac{1}{2}=2\frac{1}{2}=\frac{5}{2}\)
ngu LÊ MĨ LINH
theo thứ tự :1,6/4 =1 và 1/2,2,5/2,500
\(B=1000^9=999.1000^8+1000^8.\)
mà 999.10008 > 999.9998 = 9999 ; 10008 > 9998
\(\Rightarrow B>A\)