a, Ta có :

\(\left(\frac{1}{2}\right)^{50}=\left(\left(\frac{1}{2}\right)^5\right)^{10}=\left(\frac{1}{32}\right)^{10}\)

bạn so sánh nha :)

b,

T/c : \(99^{20}=\left(\left(99\right)^2\right)^{10}=9801^{10}\)

tiếp đây thì bạn tự làm nha có gì k hiểu ibx mk

a)

Vì 3<5

\(\Rightarrow3^{30}< 5^{30}\)

\(\Rightarrow\left(-3\right)^{30}< \left(-5\right)^{30}\)

b)

Ta có

\(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^4\right]^{10}.\left(\frac{1}{2}\right)^{10}\)

\(=\left(\frac{1}{16}\right)^{10}.\left(\frac{1}{2}\right)^{10}\)

Ta có

\(\left(\frac{1}{2}\right)^{10}< 1\)

\(\Leftrightarrow\left(\frac{1}{16}\right)^{10}.\left(\frac{1}{2}\right)^{10}< \left(\frac{1}{16}\right)^{10}\)

\(\Leftrightarrow\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{16}\right)^{10}\)

ta có :\(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)là 2 lũy thừa bậc chẵn nên :\(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)

từ trên suy ra (-5)^30<(-3)^50

b) Ta có:\(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2^5}\right)^{10}=\left(\frac{1}{32}\right)^{10}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

ta có :\(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)là 2 lũy thừa bậc chẵn nên :\(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)

từ trên suy ra (-5)^30<(-3)^50

b) Ta có:\(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2^5}\right)^{10}=\left(\frac{1}{32}\right)^{10}\)

\(\Rightarrow\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{16}\right)^{10}\)

Bài 1 và Bài 2 dễ, bn có thể tự làm được!

Bài 3:

a) ta có: 10²⁰ = (10²)¹⁰ = 100¹⁰

=> 100¹⁰>9¹⁰

=> 10²⁰>9¹⁰

b) ta có: (-5)³⁰ = 5³⁰ =( 5³)¹⁰ = 125¹⁰ ( vì là lũy thừa bậc chẵn)

(-3)⁵⁰ = 3⁵⁰ = (3⁵)¹⁰= 243¹⁰

=> 125¹⁰ < 243¹⁰

=> (-5)³⁰ < (-3)⁵⁰

c) ta có: 64⁸ = (2⁶)⁸= 2⁴⁸

16¹² = ( 2⁴)¹² = 2⁴⁸

=> 64⁸ = 16¹²

d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)

\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)

\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

3.a) Ta có: 9¹⁰=(3²)¹⁰=3²⁰

Mà 10²⁰<3²⁰

Nên 10²⁰<9¹⁰

c)Ta có:64⁸ =(8²)⁸=8¹⁶

16¹²=(2³)¹²=8³⁶

vì 8¹⁶<8³⁶

Nên 64⁸<16²

Vì\(\left(\frac{1}{16}\right)^{10}\)= \(\left[\left(\frac{1}{2}\right)^4\right]^{10}\)= \(\left(\frac{1}{2}\right)^{40}\)

Mà 40<50 =>\(\left(\frac{1}{2}\right)^{40}\)< \(\left(\frac{1}{2}\right)^{50}\)hay \(\left(\frac{1}{16}\right)^{10}\)< \(\left(\frac{1}{2}\right)^{50}\)

Vậy \(\left(\frac{1}{16}\right)^{10}\)<\(\left(\frac{1}{2}\right)^{50}\)

Học giỏi!^^ (đúng thì k cho mik nhé,cảm ơn!)

\(\left(\frac{1}{2}\right)^{50}=\left(\left(\frac{1}{2}\right)^5\right)^{10}=\left(\frac{1}{32}\right)^{10}\)
Ta có\(\frac{1}{16}>\frac{1}{32}\)nên\(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)hay\(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

Ta có:

\(16^{10}=\left(2^4\right)^{10}=2^{4\cdot10}=2^{40}< 2^{50}\)

=>\(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

Vậy ......

Bài 1:

Ta có:

\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)

\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)

Lại có:

\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)

\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)

Bài 2:

Ta có:

\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)

\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)

Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)

\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)

\(\Rightarrow13A>13B\Rightarrow A>B\)

Ta có:

\(\left(\frac{1}{16}\right)^{50}=\left[\left(\frac{1}{2}\right)^4\right]^{50}=\left(\frac{1}{2}\right)^{200}=\frac{1^{200}}{2^{200}}=\frac{1}{2^{200}}\)

\(\left(\frac{1}{2}\right)^{60}=\frac{1^{60}}{2^{60}}=\frac{1}{2^{60}}\)

Vì \(2^{200}>2^{60}\Rightarrow\frac{1}{2^{200}}< \frac{1}{2^{60}}\Rightarrow\left(\frac{1}{16}\right)^{50}< \left(\frac{1}{2}\right)^{60}\)

Ta có:

\(\left(\frac{1}{16}\right)^{50}=\left(\frac{1}{2}\right)^{4.50}=\left(\frac{1}{2}\right)^{200}\)

\(\Rightarrow\left(\frac{1}{2}\right)^{500}>\left(\frac{1}{2}\right)^{60}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{50}>\left(\frac{1}{2}\right)^{60}\)

\(\left(\frac{1}{16}\right)^{50}=\left[\left(\frac{1}{2}\right)^4\right]^{50}=\left(\frac{1}{2}\right)^{200}\)

Vì \(\frac{1}{2}=\frac{1}{2}\) mà \(200>60\)

=> \(\left(\frac{1}{2}\right)^{200}>\left(\frac{1}{2}\right)^{60}\)

=>\(\left(\frac{1}{16}\right)^{50}>\left(\frac{1}{2}\right)^{60}\)

(\(\frac{1}{2}\))⁵⁰=(\(\frac{1}{2^5}\))¹⁰=(\(\frac{1}{32}\))¹⁰

Do 1/6> 1/30 nên (\(\frac{1}{6}\))¹⁰>(\(\frac{1}{2}\))⁵⁰

\(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left[\frac{1^5}{2^5}\right]^{10}=\left[\frac{1}{32}\right]^{10}\)

Vì 2 phân số này có cùng tử mà 6 < 30 

=> \(\frac{1}{6}>\frac{1}{30}\)

=> \(\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)

Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)

\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)

Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)

               \(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)

               \(=\frac{1.2....18.19}{2.3...19.20}\)

               \(=\frac{1}{20}>\frac{1}{21}\)

Vậy A > 1/21