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

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

\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)

\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)

\(=-\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)

\(=-\frac{1.2....99}{2.3...100}.\frac{3.4....101}{2.3...100}\)

\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}\)

Học good

\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)

\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)

\(=-\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{99.101}{100^2}\)

\(=-\frac{1.2...99}{2.3...100}\cdot\frac{3.4...101}{2.3.100}\)

\(=-\frac{1}{100}\cdot\frac{101}{2}\)

\(=-\frac{101}{200}\)

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

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

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

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

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

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

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

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

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

=(-1/2) : (-2/3) :( -3/4) :...: (-49/50) 

= -1/2 . (-3/2) . (-4/3) . ... . (-50/49)

= -1/2.(-1/2) . (-50)

= - 1/100

mik ko bik viết kiểu của bn nên khó nhìn, thông cảm nha

à nhầm 

(1/2 -1) : ( 1/3-1) : ( 1/4 -1) : ... : ( 1/50 - 1) = -1/2 : -2/3 : -3/4 : ... : -49/50 

= -1/2 . (-3/2) . ( -4/3) . ... . (-50/49)

= -( 1/2.1/2 . 50)

= -25/2

= 

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 \(200>60\)

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

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

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