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ta có: \(\frac{31+32+35}{34}=\frac{31}{34}+\frac{32}{34}+\frac{35}{34}.\)
mà \(\frac{31}{32}>\frac{31}{34};\frac{32}{33}>\frac{32}{34}\)
\(\Rightarrow\frac{31}{32}+\frac{32}{33}+\frac{35}{34}>\frac{31}{34}+\frac{32}{34}+\frac{35}{34}=\frac{31+32+35}{34}\)
=00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Ta thấy : các số hạng trong tổng S đều \(>\frac{7}{35}\)
\(\Rightarrow S>\frac{7}{35}+\frac{7}{35}+\frac{7}{35}+\frac{7}{35}+\frac{7}{35}\)
\(\Rightarrow S>\frac{35}{35}\)
\(\Rightarrow S>1\) ( đpcm )
\(\frac{2015^{35}+1}{2015^{34}+1}=\frac{2015^{35}+2015-2014}{2015^{34}+1}=\frac{2015\left(2015^{34}+1\right)-2014}{2015^{34}+1}=\frac{2015\left(2015^{34}+1\right)}{2015^{34}+1}-\frac{2014}{2015^{34}+1}=2015-\frac{2014}{2015^{34}+1}\)
\(\frac{2015^{34}+1}{2015^{33}+1}=\frac{2015^{34}+2015-2014}{2015^{33}+1}=\frac{2015\left(2015^{33}+1\right)-2014}{2015^{33}+1}=\frac{2015\left(2015^{33}+1\right)}{2015^{33}+1}-\frac{2014}{2015^{33}+1}=2015-\frac{2014}{2015^{33}+1}\)
Mà \(2015-\frac{2014}{2015^{34}+1}>2015-\frac{2014}{2015^{33}+1}\)
Vậy\(\frac{2015^{35}+1}{2015^{34}+1}>\frac{2015^{34}+1}{2015^{33}+1}\)
Mình thấ bài này hơi khó sao chúng ta ko thử giải theo cách khác
34.\(\frac{34}{33}\) .35 hay\(\frac{34.34}{33.35}\)
vậy bn?
\(\frac{34x34}{33x35}=\frac{34x\left(33+1\right)}{33x\left(34+1\right)}=\frac{34x33+34}{33x34+33}=\frac{34}{33}\)
#Phoenix