a) Ta có: \(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3}\right)^{5\cdot6}=\left(\dfrac{1}{3}\right)^{30}\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{28}>\left(\dfrac{1}{243}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{3^4}\right)^7>\left(\dfrac{1}{243}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\)

mà \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7\)

nên \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)

\(\left(\dfrac{3}{8}\right)^5\&\left(\dfrac{5}{243}\right)^3\)
\(\left(\dfrac{3}{8}\right)^5=\left(\dfrac{90}{240}\right)^5=\dfrac{90^5}{240^5}\)

\(\left(\dfrac{5}{243}\right)^3=\dfrac{5^3}{243^3}\)

\(=>\dfrac{90^5}{240^5}>\dfrac{5^3}{243^3}\)

\(=>\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)

\(\left(\dfrac{1}{80}\right)^7\&\left(\dfrac{1}{243}\right)^6\)

\(\dfrac{1}{80}>\dfrac{1}{81}=\dfrac{1}{3^4}\)

\(=>\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{3^4}\right)^7=\dfrac{1}{3^{7.4}}=\dfrac{1}{3^{28}}>\dfrac{1}{3^{30}}\)

\(=\dfrac{1}{\left(3^5\right)^6}=\left(\dfrac{1}{243}\right)^6\)

\(=>\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)

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

ta có

2 mũ 3 = 8

3 mũ 2=9

vì 8<9 nên 2 mũ 3 <3 mũ 2

 Ta có: 333^444= 111^444 x 3^444 
444^333 = 111^333 x 4^333 
Tách: 3^444 = (3^4)^111 =81^111 <=>4^333 = (4^3)^111 = 64^111 
Mà: {111^444 > 111^333 (1) 
{81^111 > 64^111 hay: (3^4)^111 > (4^3)^111 (2) 
Từ (1) và (2) ta có:333^444 > 444^333 

tick cái bạn

a,444^333>333^444

b3^486>4^363

c,5^217<123^72

d,31^11>17^14

a) Ta có:

\(2^{300}=2^{3\cdot100}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=3^{2\cdot100}=\left(3^2\right)^{100}=9^{100}\)

Mà: \(8< 9\)

\(\Rightarrow8^{100}< 9^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) Ta có:

\(3^{500}=3^{5\cdot100}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=7^{3\cdot100}=\left(7^3\right)^{100}=343^{100}\)

Mà: \(243< 343\)

\(\Rightarrow243^{100}< 343^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

c) Ta có: 

\(8^5=\left(2^3\right)^5=2^{3\cdot5}=2^{15}=2\cdot2^{15}\)

\(3\cdot4^7=3\cdot\left(2^2\right)^7=3\cdot2^{2\cdot7}=3\cdot2^{14}\)

Mà: \(2< 3\)

\(\Rightarrow2\cdot2^{14}< 3\cdot2^{14}\)

\(\Rightarrow8^5< 3\cdot4^7\)

d) Ta có:

\(202^{303}=202^{3\cdot101}=\left(202^3\right)^{101}=8242408^{101}\)

\(303^{202}=303^{2\cdot101}=\left(303^2\right)^{101}=91809^{101}\)

Mà: \(8242408>91809\)

\(\Rightarrow8242408^{101}>91809^{101}\)

\(\Rightarrow202^{303}>303^{202}\)

a,5mũ 36=(5mũ3)mũ12=125 mũ12

11^24=(11^2)12=121^12

vì 121<125 nên 5^36>11^24

cảm ơn nha

bạn ơi sao bài này khó thế?

giúp minh

a: \(=32-76=-44\)

b: \(=20-30+4^2=-10+16=6\)