\(4^{127}=4\cdot4\cdot4\cdot4\cdot4\cdot...\cdot4\cdot4=2,894802231e\)

\(81^{43}=81\cdot81\cdot81\cdot81\cdot81\cdot...\cdot81\cdot81=1,16106307e\)

\(\Rightarrow4^{127}< 81^{43}\)

Ta có 4¹²⁷=(4³)⁴²*4=64⁴²*4

         81⁴³=81⁴²*81

81⁴²>64⁴²   ;    81>4

\(\Rightarrow81^{42}\cdot81>64^{42}\cdot4\Rightarrow81^{43}>4^{127}\)

4 ¹²⁷< 4¹²⁹=(4³)⁴³=64⁴³

Vi 64⁴³<81⁴³

=.> 4¹²⁷< 81⁴³

Có : 81^43 = (3^4)^43 = 3^172 < 4^2017

=> 81^43 < 4^2017 

Vậy 81^43 < 4^2017

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\(1-\frac{43}{44}\)=\(\frac{1}{44}\)

\(1-\frac{65}{66}=\frac{1}{66}\)

\(\frac{1}{44}>\frac{1}{66}\)

=> \(\frac{43}{44}>\frac{65}{66}\)

\(\dfrac{43}{87}\)<\(\dfrac{37}{73}\)

a) \(\frac{42}{-37}=-\frac{42}{37}=-\frac{37+5}{37}=-1-\frac{5}{37}\) ; \(-\frac{56}{43}=-\frac{43+13}{43}=-1-\frac{13}{43}\)

So sánh \(\frac{5}{37}\) với \(\frac{13}{43}\) thì dễ dàng được \(\frac{5}{37}< \frac{13}{43}\)

Do đó \(\frac{42}{-37}>\frac{-56}{43}\)

b) Dễ thấy \(\frac{37}{67}>0>-\frac{377}{677}\)

Bài 1. Tính căn bậc hai số học của các số sau:

1) 36=\(\sqrt{36}=4\)

2) 81\(\sqrt{81}=9\)

3) 121=\(\sqrt{121}=11\)

4) 144=\(\sqrt{144}=12\)

5) 0,16=\(\sqrt{0,16}=0,4\)

7) 29=\(\sqrt{29}~5,39\)

8) 0=\(\sqrt{0}=0\)

Bài 2: 

1: \(\sqrt{6}< \sqrt{41}\)

2: \(\sqrt{19}>\sqrt{4}\)

3: \(\sqrt{21}>\sqrt{5}\)

4: \(\sqrt{7}< \sqrt{51}\)

A) Ta có: 

\(\dfrac{12}{13}=\dfrac{13}{13}-\dfrac{1}{13}=1-\dfrac{1}{13}\)

\(\dfrac{13}{14}=\dfrac{14}{14}-\dfrac{1}{14}=1-\dfrac{1}{14}\)

Mà \(1-\dfrac{1}{13}< -\dfrac{1}{14}\)

\(\Rightarrow\dfrac{12}{13}< \dfrac{13}{14}\)

B) Ta có:

\(\dfrac{125}{251}=\dfrac{251}{251}-\dfrac{126}{251}=1-\dfrac{126}{251}\)

\(\dfrac{127}{253}=\dfrac{253}{253}-\dfrac{126}{253}=1-\dfrac{126}{253}\)

Mà: \(1-\dfrac{126}{251}< 1-\dfrac{126}{253}\)

\(\Rightarrow\dfrac{125}{251}< \dfrac{127}{253}\)