10³và 2¹⁰⁰

Ta có:10³⁰=(10³)¹⁰=1000¹⁰

          2¹⁰⁰=(2¹⁰)¹⁰=1024¹⁰

Vì 1000<1024 nên 10³⁰<2¹⁰⁰

5³⁰⁰ và 3⁴⁵³

Ta có:5³⁰⁰=(5²)¹⁵⁰=25¹⁵⁰

            3⁴⁵³=(3³)¹⁵¹=27¹⁵¹=27.27¹⁵⁰

Vì  25 < 27.27 nên 5³⁰⁰<3⁴⁵³

nhớ k ch mình nhé

1) \(A=\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)

\(B=\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)

\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}=\left(1+\sqrt{180}\right)-\sqrt{21}>0\)

\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)

2) \(C=\left(\sqrt{5}+\sqrt{10}+1\right)^2=5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)

\(=26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(\sqrt{35}\right)^2\)

Vậy \(\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)

3) \(\left(\frac{15-2\sqrt{10}}{3}\right)^2=\frac{225-60\sqrt{10}+40}{9}=\frac{265-60\sqrt{10}}{9}=\frac{265}{9}-\frac{20\sqrt{10}}{3}< 15\)

Vậy nên \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)

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, \(\frac{-17}{24}< \frac{-25}{31}\)

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

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

A)>

B)<

C)<

Giải chi tiết giúp mình

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 \(2^{30}+3^{30}+4^{30}>3.24^{10}\)

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

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

\(\text{Vậy }2^{30}< 3^{20}\)

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

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

\(\text{Vậy }5^{300}< 243^{100}\)

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

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

\(\text{Vậy ...}\)

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

quy đồng các phân số sao cho chúng cùng mẫu là so sánh được

Ta có:

a)18/91=18:91=0,197802197

   23/114=23:114=0,201754386

Mà:0,197802197<0,201754386 nên 18/91<23/114

b)21/52=21:52=0,403846153

   213/523=213:523=0,407265774

Mà:0,403846153<0,407265774 nên 21/52<213/523

c)1313/9191=1313:9191=0,142857142

   1111/7373=1111:7373=0,150684931

Mà:0,142857142<0,150684931 nên 1313/9191<1111/7373

^^^^!~~~
 

a) \(\frac{18}{91}=18\div91=0,1978021978\)

\(\frac{23}{114}=23\div114=0,20175438596\)

Mà \(0,1978021978< 0,20175438596\)nên \(\frac{18}{91}< \frac{23}{114}\) 

b) Cũng tương tự như phần a.

c) Cũng tương tự như phần a.