\(3^{200}=9^{100}>4^{100}\\ 5^{200}=25^{100}< 64^{100}=4^{300}\\ 6^{50}=36^{25}>7^{25}\\ 8^{40}=64^{20}>10^{20}\\ 16^{20}=256^{10}>32^{10}\)

tick mik nha!!

3200=9100>41005200=25100<64100=4300650=3625>725840=6420>10201620=25610>3210

a: \(3^{200}=\left(3^2\right)^{100}=9^{100}>4^{100}\)

b: \(5^{200}=\left(5^2\right)^{100}=25^{100}\)

\(4^{300}=\left(4^3\right)^{100}=64^{100}\)

mà 25<64

nên \(5^{200}< 4^{300}\)

c: \(6^{50}=\left(6^2\right)^{25}=36^{25}>7^{25}\)

https://olm.vn/hoi-dap/question/1038454.html 

Mình vừa làm cách đây 11 phút nhé !

Ta có : a²⁰¹⁰ + b²⁰¹⁰ + c²⁰¹⁰ = a¹⁰⁰⁵b¹⁰⁰⁵ + b¹⁰⁰⁵c¹⁰⁰⁵ + c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰ = 2a¹⁰⁰⁵b¹⁰⁰⁵ + 2b¹⁰⁰⁵c¹⁰⁰⁵ + 2c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰  - 2a¹⁰⁰⁵b¹⁰⁰⁵ - 2b¹⁰⁰⁵c¹⁰⁰⁵ - 2c¹⁰⁰⁵a¹⁰⁰⁵ = 0

<=> (a²⁰¹⁰ - 2a¹⁰⁰⁵b¹⁰⁰⁵ + b²⁰¹⁰) + (b²⁰¹⁰ - 2b¹⁰⁰⁵c¹⁰⁰⁵ + c²⁰¹⁰) + (c²⁰¹⁰ - 2c¹⁰⁰⁵a¹⁰⁰⁵  +  a²⁰¹⁰​) = 0

<=> (a¹⁰⁰⁵ - b¹⁰⁰⁵)² + (b¹⁰⁰⁵ - c¹⁰⁰⁵)² + (c¹⁰⁰⁵ - a¹⁰⁰⁵ ​)² = 0

=> a¹⁰⁰⁵ - b¹⁰⁰⁵ = b¹⁰⁰⁵ - c¹⁰⁰⁵ = c¹⁰⁰⁵ - a¹⁰⁰⁵ ​ = 0

=> a = b = c 

Vậy (a - b)²⁰ + (b - c)¹¹ + (c - a)²⁰¹⁰ = (a - a)²⁰ + (a - a)¹¹ + (a - a)²⁰¹⁰ = 0 + 0 + 0 = 0 .

Ta có : a²⁰¹⁰ + b²⁰¹⁰ + c²⁰¹⁰ = a¹⁰⁰⁵b¹⁰⁰⁵ + b¹⁰⁰⁵c¹⁰⁰⁵ + c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰ = 2a¹⁰⁰⁵b¹⁰⁰⁵ + 2b¹⁰⁰⁵c¹⁰⁰⁵ + 2c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰  - 2a¹⁰⁰⁵b¹⁰⁰⁵ - 2b¹⁰⁰⁵c¹⁰⁰⁵ - 2c¹⁰⁰⁵a¹⁰⁰⁵ = 0

<=> (a²⁰¹⁰ - 2a¹⁰⁰⁵b¹⁰⁰⁵ + b²⁰¹⁰) + (b²⁰¹⁰ - 2b¹⁰⁰⁵c¹⁰⁰⁵ + c²⁰¹⁰) + (c²⁰¹⁰ - 2c¹⁰⁰⁵a¹⁰⁰⁵  +  a²⁰¹⁰​) = 0

<=> (a¹⁰⁰⁵ - b¹⁰⁰⁵)² + (b¹⁰⁰⁵ - c¹⁰⁰⁵)² + (c¹⁰⁰⁵ - a¹⁰⁰⁵ ​)² = 0

=> a¹⁰⁰⁵ - b¹⁰⁰⁵ = b¹⁰⁰⁵ - c¹⁰⁰⁵ = c¹⁰⁰⁵ - a¹⁰⁰⁵ ​ = 0

=> a = b = c 

Vậy (a - b)²⁰ + (b - c)¹¹ + (c - a)²⁰¹⁰

 = (a - a)²⁰ + (a - a)¹¹ + (a - a)²⁰¹⁰ 

= 0 + 0 + 0

= 0 .

=> ĐPCM

Ta có : a²⁰¹⁰ + b²⁰¹⁰ + c²⁰¹⁰ = a¹⁰⁰⁵b¹⁰⁰⁵ + b¹⁰⁰⁵c¹⁰⁰⁵ + c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰ = 2a¹⁰⁰⁵b¹⁰⁰⁵ + 2b¹⁰⁰⁵c¹⁰⁰⁵ + 2c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰  - 2a¹⁰⁰⁵b¹⁰⁰⁵ - 2b¹⁰⁰⁵c¹⁰⁰⁵ - 2c¹⁰⁰⁵a¹⁰⁰⁵ = 0

<=> (a²⁰¹⁰ - 2a¹⁰⁰⁵b¹⁰⁰⁵ + b²⁰¹⁰) + (b²⁰¹⁰ - 2b¹⁰⁰⁵c¹⁰⁰⁵ + c²⁰¹⁰) + (c²⁰¹⁰ - 2c¹⁰⁰⁵a¹⁰⁰⁵  +  a²⁰¹⁰​) = 0

<=> (a¹⁰⁰⁵ - b¹⁰⁰⁵)² + (b¹⁰⁰⁵ - c¹⁰⁰⁵)² + (c¹⁰⁰⁵ - a¹⁰⁰⁵ ​)² = 0

=> a¹⁰⁰⁵ - b¹⁰⁰⁵ = b¹⁰⁰⁵ - c¹⁰⁰⁵ = c¹⁰⁰⁵ - a¹⁰⁰⁵ ​ = 0

=> a = b = c 

Vậy (a - b)²⁰ + (b - c)¹¹ + (c - a)²⁰¹⁰ = (a - a)²⁰ + (a - a)¹¹ + (a - a)²⁰¹⁰ = 0 + 0 + 0 = 0 .

       a²⁰¹⁰ + b²⁰¹⁰ + c²⁰¹⁰ = a¹⁰⁰⁵b¹⁰⁰⁵ + b¹⁰⁰⁵c¹⁰⁰⁵ + c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰ = 2a¹⁰⁰⁵b¹⁰⁰⁵ + 2b¹⁰⁰⁵c¹⁰⁰⁵ + 2c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰  - 2a¹⁰⁰⁵b¹⁰⁰⁵ - 2b¹⁰⁰⁵c¹⁰⁰⁵ - 2c¹⁰⁰⁵a¹⁰⁰⁵ = 0

<=> (a²⁰¹⁰ - 2a¹⁰⁰⁵b¹⁰⁰⁵ + b²⁰¹⁰) + (b²⁰¹⁰ - 2b¹⁰⁰⁵c¹⁰⁰⁵ + c²⁰¹⁰) + (c²⁰¹⁰ - 2c¹⁰⁰⁵a¹⁰⁰⁵  +  a²⁰¹⁰​) = 0

<=> (a¹⁰⁰⁵ - b¹⁰⁰⁵)² + (b¹⁰⁰⁵ - c¹⁰⁰⁵)² + (c¹⁰⁰⁵ - a¹⁰⁰⁵ ​)² = 0

=> a¹⁰⁰⁵ - b¹⁰⁰⁵ = b¹⁰⁰⁵ - c¹⁰⁰⁵ = c¹⁰⁰⁵ - a¹⁰⁰⁵ ​ = 0

=> a = b = c 

Ta có \(\left(a^{1005}-b^{1005}\right)^2+\left(b^{1005}-c^{1005}\right)^2+\left(c^{1005}-a^{1005}\right)^2>0\Leftrightarrow a^{2010}-2a^{1005}b^{1005}+b^{2010}+b^{2010}-2b^{1005}c^{1005}+c^{2010}+c^{2010}-2a^{1005}c^{1005}+a^{1005}>0\Leftrightarrow2\left(a^{2010}+b^{2010}+c^{2010}\right)-2\left(a^{1005}b^{1005}+a^{1005}c^{1005}+c^{1005}b^{1005}\right)>0\Leftrightarrow a^{2010}+b^{2010}+c^{2010}>a^{1005}b^{1005}+a^{1005}c^{1005}+c^{1005}b^{1005}\)(đpcm)

0

10^1005 > 9^1005 = 3^2.1005 = 3^2010 

=> 10^1005 > 3^2010 

Chúc bn học giỏi nha ^_^

3^2010>10^1005

3^2010>10^1005

chúc bạn học tốt nha 

k mk nha bạn

Đặt \(\left\{{}\begin{matrix}a^{1005}=x\\b^{1005}=y\\c^{1005}=z\end{matrix}\right.\) \(\Rightarrow x^2+y^2+z^2=xz+xz+yz\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2xz+2yz\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-z=0\\y-z=0\end{matrix}\right.\) \(\Leftrightarrow x=y=z\)

\(\Rightarrow a^{1005}=b^{1005}=c^{1005}\Rightarrow a=b=c\)

\(\Rightarrow M=0\)

bạn cũng xem phim gia sư siêu quậy reborn à ?

xem à ?

\(3\sqrt{3}-2\sqrt{2}=\sqrt{27}-\sqrt{8}>\sqrt{25}-\sqrt{9}=5-3\)

\(\Rightarrow3\sqrt{3}-2\sqrt{2}>2\)