\(23^{32}\) và \(32^{23}\)

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25 tháng 5 2020


Ta có: \(\frac{23}{32}=1-\frac{9}{32}\)

          \(\frac{39}{48}=1-\frac{9}{48}\)

Vì \(\frac{9}{32}>\frac{9}{48}\) => \(\frac{23}{32}< \frac{39}{48}\)

Vậy ........

15 tháng 7 2017

a/ \(8^5=\left(2^3\right)^5=2^{15}\)và \(32^3=\left(2^5\right)^3=2^{15}\Rightarrow8^5=32^3\)

b/ \(27^4=\left(3^3\right)^4=3^{12}\) và \(9^6=\left(3^2\right)^6=3^{12}\Rightarrow27^4=9^6\)

c/ \(23^{17}-23^{16}=23^{16}\left(23-1\right)=22.23^{16}\)

\(23^{16}-23^{15}=23^{15}\left(23-1\right)=22.23^{15}\)

\(\Rightarrow22.23^{16}>22.23^{15}\Rightarrow23^{17}-23^{16}>23^{16}-23^{15}\)

d/ \(\frac{3^{2015}+1}{3^{2016}}=\frac{1}{3}+\frac{1}{3^{2016}}\) và \(\frac{3^{2016}+1}{3^{2017}+1}=\frac{3^{2017}+3}{3\left(3^{2017}+1\right)}=\frac{3^{2017}+1+2}{3\left(3^{2017}+1\right)}=\frac{1}{3}+\frac{2}{3}.\frac{1}{3^{2017}+1}\)

\(\frac{1}{3^{2016}}>\frac{1}{3^{2017}}>\frac{1}{3^{2017}+1}>\frac{2}{3}.\frac{1}{3^{2017}+1}\)

\(\Rightarrow\frac{3^{2015}+1}{3^{2016}}>\frac{3^{2016}+1}{3^{2017}+1}\)

Câu cuối phân tích tương tự

15 tháng 8 2015

a)\(\frac{-23}{47}\)\(\frac{-31}{61}\)

= -23x61= -1403

=-31x47= -1457

vậy là \(\frac{-23}{47}\)>\(\frac{-31}{61}\)

vì -1403 > -1457

câu b) cũng như vậy

23 tháng 4 2015

\(49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

\(=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

\(=\left(49-14\right)+\left(\frac{8}{23}-\frac{8}{23}\right)+5\frac{7}{32}\)

\(=35+5\frac{7}{32}\)

\(=34\frac{32}{32}+5\frac{7}{32}\)

\(=\left(34+5\right)+\left(\frac{32}{32}+\frac{7}{32}\right)\)

\(=39+\frac{39}{32}\)

\(=39\frac{39}{32}=40\frac{7}{32}\)

29 tháng 3 2019

\(B=\frac{23^{41}+1}{23^{42}+1}\)

Vì B < 1

\(\Rightarrow B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23(23^{40}+1)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

P/s: Hoq chắc

29 tháng 3 2019

ta có 

\(B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23\left(23^{40}+1\right)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

\(\Rightarrow B< A\)