\(S=1+2+2^2+2^3+...+2^9\) 

Đặt \(2S=2+2^2+2^3+2^4+...+2^{10}\) 

\(2S-S=2^{10}-1\) hay \(S=2^{10}-1< 2^{10}\)

\(\Rightarrow\) \(2^{10}=2^2.2^8< 5.2^8\) 

Vậy \(S< 5.2^8\)

\(#Tuyết\)

2S=2+2^2+...+2^10

=>S=2^10-1=1023

5*2^8=256*5=1280

=>S<5*2^8

`@` `\text {Answer}`

`\downarrow`

`S = 1 + 2 + 2^2 + 2^3 + ... + 2^9`

`=> 2S = 2 + 2^2 + 2^3 + ... + 2^10`

`=> 2S - S = (2+2^2 + 2^3 + ... + 2^10) - (1 + 2 + 2^2 + 2^3+...+2^9)`

`=> S = 2^10 - 1`

Mà `2^10 - 1 < 2^10`

`=> S < 2^10 (1)`

Ta có:

 `2^10 = 2^7*8`

Mà `5*2^8 = 5* 2 * 2^7 = 10* 2^7`

Vì `10 > 8 => 2^7 * 8 < 2^7  * 10 (2)`

Từ `(1)` và `(2)`

`=> S < 5 * 2^7``.`

a: B=31/39=62/78=1-16/78; A=35/37=280/296=1-16/296

16/78>16/296

=>B<A

b: C=23/27=46/54=1-8/54

D=1-8/29

8/29>8/54

=>D<C

a)

Có: 

\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)

Vì \(\sqrt{117}>\sqrt{116}\)  nên \(3\sqrt{13}>2\sqrt{29}\)

b)

Có:

\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)

\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)

Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\)  nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)

c)

Có:

\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)

\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)

Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)

d)

Có:

\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)

\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)

lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)

 \(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)

23/27> 22/29

23/27>22/29

a) `14/21=(14:7)/(21:7)=2/3=4/6`

`60/72=(60:12)/(72:12)=5/6`

Vì `4/6 <5/6`

`=> 14/21 < 60/72`

b) `22/37 = (22:2)/(37:2)= 11/(37/2)`

Vì `54 > 37/2`

`=> 11/54 < 22/37` 

a: \(\left(4+\sqrt{33}\right)^2=49+8\sqrt{33}=49+2\cdot\sqrt{528}\)

\(\left(\sqrt{29}+\sqrt{14}\right)^2=43+2\cdot\sqrt{29\cdot14}=43+2\cdot\sqrt{406}\)

mà 49>43 và 528>406

nên \(\left(4+\sqrt{33}\right)^2>\left(\sqrt{29}+\sqrt{14}\right)^2\)

=>\(4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)

a) \(\dfrac{-1}{20}=\dfrac{-7}{140}\)

\(\dfrac{5}{7}=\dfrac{100}{140}\)

mà -7<100

nên \(-\dfrac{1}{20}< \dfrac{5}{7}\)

b) \(\dfrac{216}{217}< 1\)

\(1< \dfrac{1164}{1163}\)

nên \(\dfrac{216}{217}< \dfrac{1164}{1163}\)

c) \(\dfrac{-12}{17}=\dfrac{-180}{255}\)

\(\dfrac{-14}{15}=\dfrac{-238}{255}\)

mà -180>-238

nên \(-\dfrac{12}{17}>\dfrac{-14}{15}\)

d) \(\dfrac{27}{29}>0\)

\(0>-\dfrac{2727}{2929}\)

nên \(\dfrac{27}{29}>-\dfrac{2727}{2929}\)

Ta lấy ps trung gian: 23/29

So sánh : 23/27>23/29>22/29.

=> 23/27>22/29

> va

Sry > và > nha

a, \(\dfrac{23}{27}\) > \(\dfrac{22}{27}\) > \(\dfrac{22}{29}\)

Vậy \(\dfrac{23}{27}\) > \(\dfrac{22}{29}\)

b, \(\dfrac{25}{74}\) > \(\dfrac{24}{74}\) = \(\dfrac{12}{37}\)

Vậy \(\dfrac{25}{74}\) > \(\dfrac{12}{37}\)

a) Ta có: \(\frac{-13}{38}\)> \(\frac{-13}{88}\)(hai phân số cùng tử)

Lại có \(\frac{-13}{88}\)> \(\frac{-29}{88}\)(hai phân số cùng mẫu)

Suy ra: \(\frac{-13}{38}>\frac{-29}{88}\)

b) Tương tự, ta có \(\frac{22}{29}< \frac{22}{27}< \frac{24}{27}\)

\(\Rightarrow\frac{22}{29}< \frac{24}{27}\)

c)Tương tự, ta có: \(\frac{23}{29}< \frac{23}{27}< \frac{24}{27}\)

\(\Rightarrow\frac{23}{29}< \frac{24}{27}\)

d) Tương tự, ta có: \(\frac{-13}{91}>\frac{-13}{202}>\frac{-29}{202}\)

\(\Rightarrow\frac{-13}{92}>\frac{-29}{202}\)

Ps: Mình làm theo cách so sánh thông qua phân số trung gian, rất mong được tham khảo cách khác nhanh hơn!!!

\(-\frac{13}{38}>-\frac{29}{38}\)