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

Mà ta có

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

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

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

Ta có :

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

\(\Rightarrow\frac{1}{243^9}>\frac{1}{83^{13}}\)

mình chắc chắn luôn

-https://olm.vn/hoi-dap/detail/77727486175.html

(100^99+99^100)^100

(100^100+99^100)^99

ta có : (100^99+99^100)^100=100^9900+99^10000

           (100^100+99^100)^99=100^9900+99^9900

=)100^9900=100^9900; 99^10000>99^9900(vì 10000>9900)

=)(100^99+99^100)^100>(100^100+99^100)^99

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

bạn nào làm được thì giúp mình với còn bài này thì mình không biết làm. sorry nha

AI NÓI TỚ NÓI SAI, CÓ NÓI VỀ BÀI ĐÂU MÀ SAI ĐIÊN À MẤY BẠN KIA

ta có:\(\left(\dfrac{-1}{16}\right)^{10}=\left(\dfrac{1}{16}\right)^{10}=\left(\dfrac{1^4}{2^4}\right)^{10}=\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{40}=\dfrac{1^{40}}{12^{40}}=\dfrac{1}{2^{40}}\)

ta có:

\(\left(\dfrac{-1}{2}\right)^{500}=\left(\dfrac{1}{2}\right)^{500}=\dfrac{1^{500}}{2^{500}}=\dfrac{1}{2^{500}}\)

Vì 40<500

⇒2\(^{40}< 2^{500}\)

⇒\(\dfrac{1}{2^{40}}>\dfrac{1}{2^{500}}\)

⇒\(\left(\dfrac{-1}{16}\right)^{10}>\left(\dfrac{-1}{2}\right)^{500}\)

Vậy \(\left(\dfrac{-1}{16}\right)^{10}>\left(\dfrac{-1}{2}\right)^{500}\)

\(+,\left(\dfrac{-1}{16}\right)^{10}=\left(\dfrac{\left(-1\right)^4}{2^4}\right)^{10}=\left[\left(\dfrac{-1}{2}\right)^4\right]^{10}=\left(\dfrac{-1}{2}\right)^{40}\)

Vì 40<500→\(\left(\dfrac{-1}{2}\right)^{40}< \left(\dfrac{-1}{2}\right)^{500}hay\left(\dfrac{-1}{16}\right)^{10}< \left(\dfrac{-1}{2}\right)^{500}\)

a.(3^2+4^2).x=10^2

(9+16).x     =100

25.x           =100

x                =100:25

x                =4

b.(x-5)^2       =81

x-5             =9

x                =9+5

x                =14

c.(2x+1)^3 = 343

2x+1          = 7

2x              =7-1

2x              =6

x                =6:2

x                = 3

(9+16).x=100

25.x=100

x=100:25

x=4