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Ta có : \(17^{14}>16^{14}=\left(2^4\right)^{14}\)\(=2^{64}>2^{55}=\left(2^5\right)^{11}=31^{11}\)
\(\Rightarrow\)\(17^{14}>31^{11}\)
a,\(\frac{31}{17}+\frac{5}{13}+\frac{8}{13}-\frac{14}{17}=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{5}{13}+\frac{8}{13}\right)=1+1=2\) b,\(\frac{5}{7}.\frac{22}{11}+\frac{5}{7}.\frac{9}{11}+\frac{5}{7}=\frac{5}{7}.\left(\frac{22}{11}+\frac{9}{11}+1\right)=\frac{5}{7}.4=\frac{20}{7}\)
ta co 3mu11=3.3.3.3.3.3.3.3.3.3.3=177147
17mu14=17.17.17.17.17.17.17.17.17.17.17.17.17.17=......
vay 17mu14 > 3mu11
\(\frac{31}{53}.\frac{17}{28}+\frac{31}{28}.\frac{11}{53}\)
\(=\frac{31}{53}.\frac{17}{28}+\frac{11}{28}.\frac{31}{53}\)
\(=\frac{31}{53}.\left(\frac{17}{28}+\frac{11}{28}\right)\)
\(=\frac{31}{53}.1\)
\(=\frac{31}{53}\)
\(\frac{31}{53}\times\frac{17}{28}+\frac{31}{28}\times\frac{11}{53}\)
\(=\frac{31}{53}\times\frac{17}{28}+\frac{31}{53}\times\frac{11}{28}\)
\(=\frac{31}{53}\times\left(\frac{17}{28}+\frac{11}{28}\right)\)
\(=\frac{31}{53}\times1\)
\(=\frac{31}{53}\)
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
3111 < 3211 = (25)11 = 255
1714 > 1614 = (24)14 = 256
Vì 255 < 256 nên 3111 < 1714
Đáp án:
17 mũ 14 > 31 mũ 11
Giải thích các bước giải:
ta có :
31 mũ 11 < 32 mũ 11 = ( 2 mũ 5 ) mũ 11 = 2 mũ 55
17 mũ 14 > 16 mũ 14 = ( 2 mũ 4 ) mũ 14 = 2 mũ 56
ta thấy 2 mũ 56 > 2 mũ 55 nên 17 mũ 14 > 31 mũ 11