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

Cách1:Ta có:\(\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{2}\right)^{40}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{16}\right)^{10}\)

Vậy..................

Cách 2:Ta có:\(\left(\frac{1}{16}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}>\left(\frac{1}{2}\right)^{50}\)

Vậy......................

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

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

Do 2⁵⁰ > 2⁴⁰ => \(\frac{1}{2^{40}}>\frac{1}{2^{50}}\)

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

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

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

Do \(\frac{1}{6}>\frac{1}{32}\Rightarrow\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)

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

a) \(10^{20}\) và \(9^{10}\)

Vì 10 > 9 ; 20 > 10

nên \(10^{20}>9^{10}\)

Vậy \(10^{20}>9^{10}\)

b) \(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)

Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

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

Vì 243 > 125 nên \(125^{10}< 243^{10}\)

Vậy \(\left(-5\right)^{30}< \left(-3\right)^{50}\)

c) \(64^8\) và \(16^{12}\)

Ta có: \(64^8=\left(4^3\right)^8=4^{24}\)

          \(16^{12}=\left(4^2\right)^{12}=4^{24}\)

Vậy \(64^8=16^{12}\left(=4^{24}\right)\)

d) \(\left(\frac{1}{6}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)

Ta có: \(\left(\frac{1}{6}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)

Vì 40 < 50 nên \(\left(\frac{1}{2}\right)^{40}< \left(\frac{1}{2}\right)^{50}\)

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

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) \(\left(-5\right)^2\).\(\left(-5\right)^3\) = -3125

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:

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

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

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:

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

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

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