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

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

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 \(2^{40}< 2^{50}\)

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

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

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

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

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

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

\(P=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{2499}{2500}\)

\(P=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)

\(P=\frac{\left(1.2.3...49\right)\left(3.4.5...51\right)}{\left(2.3.4...50\right)\left(2.3.4...50\right)}\)

\(P=\frac{1.51}{50.2}\)

\(P=\frac{51}{100}>\frac{1}{2}\)

Kết luận: \(P>\frac{1}{2}\)

\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{400}-1\right)\)

\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{400}\right)\)

\(-A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{399}{400}\)

\(-A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot...\cdot\frac{19.21}{20.20}\)

\(-A=\frac{1\cdot2\cdot3\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\cdot\frac{3\cdot4\cdot5\cdot...\cdot21}{2\cdot3\cdot4\cdot...\cdot20}\)

\(-A=\frac{1}{20}\cdot\frac{21}{2}=\frac{21}{40}>\frac{20}{40}=\frac{1}{2}\)

\(-A>\frac{1}{2}\Rightarrow A< \frac{1}{2}\)

(\(\frac{1}{2}\))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 

=> \(\frac{1}{6}>\frac{1}{30}\)

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