\(\left(\frac{1}{2}\right)^{40}=\left(\frac{1}{2}\right)^{10\cdot4}=\left(\frac{1}{16}\right)^{10}\)

Mà ta có

\(\left(\frac{1}{32}\right)^{10}< \left(\frac{1}{16}\right)^{10}\)

\(\Rightarrow\left(\frac{1}{2}\right)^{40}>\left(\frac{1}{32}\right)^{10}\)

Ta có : 

\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)

\(\Rightarrow\frac{1}{243^9}>\frac{1}{8^{13}}\)

a) Ta có :

\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)

\(\Rightarrow27^{27}>243^{13}\)

\(\Rightarrow-27^{27}< -243^{13}\)

\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)

b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)

c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)

\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)

\(\Rightarrow4^{50}>8^{30}\)

d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)

\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)

a) Ta có :

2727>2726=(272)13=72913>243132727>2726=(272)13=72913>24313

⇒2727>24313⇒2727>24313

⇒−2727<−24313⇒−2727<−24313

⇒(−27)27<(−243)13⇒(−27)27<(−243)13

b) (18)25>(18)26=(182)13=(164)13>(1128)13(81​)25>(81​)26=(821​)13=(641​)13>(1281​)13

⇒(18)25>(1128)13⇒(81​)25>(1281​)13

⇒(−18)25<(−1128)13⇒(−81​)25<(−1281​)13

c) 450=(45)10=102410450=(45)10=102410

830=(83)10=51210<102410830=(83)10=51210<102410

⇒450>830⇒450>830

d) (19)17<(19)12<(127)12(91​)17<(91​)12<(271​)12

⇒(19)17<(127)12⇒(91​)17<(271​)12

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

1/12>1/32

=>(1/12)^10>(1/32)^10

=>(1/12)^10>(1/2)^50

Có: \(\left(\dfrac{1}{12}\right)^{10}=\dfrac{1}{12^{10}}\)

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

Do \(12< 32\Rightarrow12^{10}< 32^{10}\)

\(\Rightarrow\dfrac{1}{12^{10}}>\dfrac{1}{32^{10}}\) hay \(\left(\dfrac{1}{12}\right)^{10}>\left(\dfrac{1}{2}\right)^{50}\)

1.a)A = (1 - 1/3)(1-2/5)...(1-5/5)....(1-9/5)

      =(1-1/3)....0.....(1-9/5)

      =0

     =>đpcm.

b)ta xét:

1/2² = 1/2x2 < 1/1x2

.............

1/8² = 1/8x8 <1/7x8

=>B < 1/1x2 + 1/2x3 ... + 1 + 1/7x8

<=> B <1 - 1/2 + 1/2  - 1/3  + ... + 1/7 - 1/8

<=> B < 1 - 1/8 = 7/8 < 1

=> B < 1 => đpcm

2.a) Đặt m = 2007(2006+2007) = 2006(2006 + 2007) + (2006+2007)

      Đặt n = 2006(2007+2008) = 2006(2006+2007) + (2006 + 2006)

Ta thấy : (2006+2007) > (2006 + 2006) => m > n , áp dụng công thức "a.d > c.d <=> a/b > b/d (a,c thuộc Z// b,d thuộc N)

=> A > B

   b)ta có: D = 196 + 197/197 + 198 = (196/197+198) + (197/197+198) < 196/197 + 197/198 = C

=> C > D

c)gọi 20¹⁰ là a

ta thấy : (a + 1)(a-3) = (a - 1)(a - 3) + 2(a - 3) < (a - 1)(a - 3) + 2(a - 1) = (a - 1)(a - 1)

áp dụng: ad > bc <=> a/b > c/d ( a,b,c,d thuộc Z// b,d > 0)

=> E > F

2: \(\left(\dfrac{1}{6}\right)^{10}=\dfrac{1}{6^{10}}\)

(1/2)^50=1/2^50

mà 6^10<2^50

nên (1/6)^10<(1/2)^50

3: 30^-50=1/30^50=(1/3)^500=(1/243)^100

5^-30=1/5^30=(1/125)^100

=>30^50<5^-30

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn.

\(\left(\dfrac{3}{5}\right)^{10}\cdot\left(\dfrac{5}{3}\right)^{10}-\dfrac{13^4}{39^4}+2014^0\)

=1-(1/3)^4+1

=4/3-1/81

=107/81