c) Đặt \(A=2^0+2^1+2^2+...+2^{50}\)

\(\Leftrightarrow2A=2^1+2^2+2^3...+2^{51}\)

\(\Leftrightarrow2A-A=2^1+2^2+2^3...+2^{51}\)\(-2^0-2^1-2^2-...-2^{50}\)

\(\Leftrightarrow A=2^{51}-2^0=2^{51}-1< 2^{51}\)

Vậy \(2^0+2^1+2^2+...+2^{50}< 2^{51}\)

a)Ta có: \(\hept{\begin{cases}2^{30}=\left(2^3\right)^{10}=8^{10}\\3^{30}=\left(3^3\right)^{10}=27^{10}\\4^{30}=\left(2^2\right)^{30}=2^{60}\end{cases}}\)và \(\hept{\begin{cases}3^{20}=\left(3^2\right)^{10}=9^{10}\\6^{20}=\left(6^2\right)^{10}=36^{10}\\8^{20}=\left(2^3\right)^{20}=2^{60}\end{cases}}\)

Mà \(8^{10}< 9^{10}\); \(27^{10}< 36^{10}\);\(2^{60}=2^{60}\)nên

\(8^{10}+27^{10}+2^{60}< 9^{10}+36^{10}+2^{60}\)

hay \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)

b) Ta có: \(4^{30}=2^{30}.2^{30}=8^{10}.4^{15}\)

\(3.24^{10}=3.8^{10}.3^{10}=3^{11}.8^{10}\)

Vì \(4^{15}>3^{11}\) nên \(8^{10}.4^{15}>3^{11}.8^{10}\)

hay \(2^{30}+3^{30}+4^{30}>3.24^{10}\)

\(2^{27}=2^{3.9}=8^9\)

\(3^{18}=3^{2.9}=9^9\)

Vì \(9^9>8^9\Rightarrow3^{18}>2^{27}\)

MK chỉ làm đc câu a) thui nha :

2^27 = 2^ 3.9 = 8^9

3^18 = 3^2.9=9^9

Vì 9^9 > 8^9 => 2^27 < 2 ^18 

\(\text{a, }2^{30}=8^{10}\)

     \(\text{ }3^{20}=\left(3^2\right)^{10}=9^{10}\)

\(\text{Vậy }2^{30}< 3^{20}\)

\(\text{b, }5^{300}=\left(5^3\right)^{100}=125^{100}\)

     \(3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(\text{Vậy }5^{300}< 243^{100}\)

\(\text{c, }2^{24}=\left(2^3\right)^8=8^8\)

     \(3^{16}=\left(3^2\right)^8=9^8\)

\(\text{Vậy ...}\)

a) Ta có \(\hept{\begin{cases}2^{24}=\left(2^6\right)^4=64^4\\3^{16}=\left(3^4\right)^4=81^4\end{cases}}\)

Mà \(64< 81\)

\(\Rightarrow64^4< 81^4\)

\(\Rightarrow2^{24}< 3^{16}\)

b) Ta có \(\hept{\begin{cases}2^{300}=\left(2^3\right)^{100}=8^{100}\\3^{200}=\left(3^2\right)^{100}=9^{100}\end{cases}}\)

Mà 8 < 9  

\(\Rightarrow8^{100}< 9^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

c) Ta có \(7^{20}=\left(7^4\right)^5=2401^5\)

Ta có 71 < 2401 

\(\Rightarrow71^5< 2401^5\)

\(\Rightarrow71^5< 7^{20}\)

!! K chắc câu c

@@ Học tốt

Chiyuki Fujito

a) \(2^{24}=\left(2^3\right)^8=8^8\)

\(3^{16}=\left(3^2\right)^8=9^8\)

Ta thấy 8<9\(\Rightarrow8^8< 9^8\Rightarrow2^{24}< 3^{16}\)

b) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Thấy \(8< 9\Rightarrow8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)

c) \(7^{20}=\left(7^4\right)^5=2401^5\)

Ta thấy \(71< 2401\Rightarrow71^5< 2401^5\Rightarrow71^5< 7^{20}\)

a) \(\frac{3}{-4}=\frac{-3}{4};\frac{-1}{-4}=\frac{1}{4}\)

Vì - 3 < 1 nên \(\frac{-3}{4}< \frac{1}{4}\)

hay \(\frac{3}{-4}< \frac{-1}{-4}\)

Quy đồng mẫu ta được:

15/17=15.27/17.27=405/459

25/27=25.17/27.27=425/459

⇒405/459<425/459⇒15/17<25/27

c) quy đồng

-9/6=-36/24

6/-4=36/-24

tương đương -36/24=-36/24

suy ra -9/6=6/-4

- Mình dùng cách lớp 8 để làm câu b được không :)?

- Tham khảo câu b:

https://olm.vn/hoi-dap/tim-kiem?q=+++++++++++A=2006%5E2005+1/2006%5E2006+1B=2006%5E2006+1/2006%5E2007+1so+s%C3%A1nh+A+v%C3%A0+B&id=520258

:< dễ lém á , mng giúp mìn ii 

Xin lỗ nhé thừa số 4 bé ở câu a

\(a,\sqrt{2}+\sqrt{11}< \sqrt{3}+\sqrt{16}=\sqrt{3}+4\)

a)-17/24 > -25/31

b)-27/38 < -125/195

c)-22/111> -27/134

nhớ k nha!!!!!!!!!!!!!!!!!!

Ban co the ghi cách giải ko

a)Ta có :\(3^{60}=\left(3^3\right)^{20}=27^{20}\)

\(2^{80}=\left(2^4\right)^{20}=16^{20}\)

Mà \(27^{20}>16^{20}\Leftrightarrow3^{60}>2^{80}\)

b)Ta có :\(5^{20}=\left(5^4\right)^5=625^5\)

\(4^{25}=\left(4^5\right)^5=1024^5\)

Mà \(1024^5>625^5\Leftrightarrow5^{20}< 4^{25}\)

a, \(\frac{-17}{24}< \frac{-25}{31}\)

b,\(\frac{-27}{38}< \frac{-125}{195}\)

c,\(\frac{-22}{111}>\frac{-27}{134}\)

A)>

B)<

C)<

Giải chi tiết giúp mình