\(\Leftrightarrow2\left(a^{2010}+b^{2010}+c^{2010}\right)=2\left(a^{1005}b^{1005}+b^{1005}c^{1005}+c^{1005}a^{1005}\right)\)

\(\Leftrightarrow2a^{2010}+2b^{2010}+2c^{2010}-2a^{1005}b^{1005}-2b^{1005}c^{1005}-2c^{1005}a^{1005}=0\)

\(\Leftrightarrow\left(a^{2010}-2a^{1005}b^{1005}+b^{2010}\right)+\left(b^{2010}-2b^{1005}c^{1005}+c^{2010}\right)+\left(c^{2010}-2c^{1005}a^{1005}+a^{2010}\right)=0\)

\(\Leftrightarrow\left(a^{1005}-b^{1005}\right)^2+\left(b^{1005}-c^{1005}\right)^2+\left(c^{1005}-a^{1005}\right)^2=0\)

\(\Rightarrow\left(a^{1005}-b^{1005}\right)^2=0;\left(b^{1005}-c^{1005}\right)^2=0;\left(c^{1005}-a^{1005}\right)^2=0\)

\(\Rightarrow a=b=c\)

\(\Rightarrow\left(a-a\right)^{20}+\left(a-a\right)^{11}+\left(a-a\right)^{2010}=0\)

2 ( a trên 2010 + b trân 2010 + c trên 2010 ) = 2 ( a trên 1005 b trên 1005 + b trên 1005 c trên 1005 + c trên 1005 a trên 1005 )

2a^ ( 2010 ) + 2b^ ( 2010 ) + 2c^ ( 2010 ) - 2a^ ( 1005 ) b^ ( 1005 ) - 2b^ ( 1005 ) c^ ( 1005 ) - 2c^ ( 1005 )a^ ( 1005 ) = O\)

( a^ ( 2010 ) - 2a^ ( 1005 ) b^ ( 1005 ) + b^ ( 2010 ) + ( b^( 2010 ) - 2b^ ( 1005 ) c^ ( 1005 ) + c^ ( 2010 ) + ( c^ ( 2010 ) - 2c^ ( 1005 ) a^ ( 1005 ) + a^ ( 2010 ) = 0\)

( a^ ( 1005 ) ^2 + ( b^ ( 1005 ) - c^ ( 1005 ) ^2 + ( c^ ( 1005 ) - a^ ( 1005 ) - a^ ( 1005 ) ^2 = 0\)

( a^ ( 1005 ) - b^ ( 1005 ) ^ 2= 0 : ( b^ ( 1005 ) - c^ ( 1005 ) ^2 = 0 : ( c^ ( 1005 ) - a^ ( 1005 ) ^2 = 0\)

a = b = c

( a - a ) ^ ( 20 ) + ( a - a ) ^ ( 11 ) + ( a - a ) ^ (2010 = 0\)

Vậy :  ( a -a ) ^ ( 20 ) + ( a - a ) ^ ( 11 ) + ( a + a ) ^ ( 2010 = 0\)

1 ) 2(ax2010+bx2010+cx2010)=2(ax1005bx1005+bx1005cx1005+cx1005ax1005)

2ax2010+2bx2010+2cx2010−2ax1005bx1005−2bx1005cx1005−2cx1005ax1005=0

(a2010−2a1005b1005+b2010)+(b2010−2b1005c1005+c2010)+(c2010−2c1005a1005+a2010)=0

Suy ra a=b=c=0

Vậy biểu thức có kết quả là 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 .

       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 

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

\(a^{2010}+b^{2010}+c^{2010}=a^{1005}b^{1005}+b^{1005}c^{1005}+a^{1005}c^{1005}\)

=>\(2a^{2010}+2b^{2010}+2c^{2010}-2a^{1005}b^{1005}-2b^{1005}c^{1005}-2a^{1005}c^{1005=0}\)

=>\(\left(a^{1005}-b^{1005}\right)\left(b^{1005}-c^{1005}\right)\left(a^{1005}-c^{1005}\right)=0\)

=>a=b=c

\(A=\left(b-b\right)^{20}+\left(b-b\right)^{11}+\left(c-c\right)^{2010}=0\)

Ta có : \(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)(bunhiacopxki)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge2010^2\)

\(\Rightarrow P=a^2+b^2+c^2\ge\frac{2010^2}{3}=1346700\)

tại sao lại là 3(a^2+b^2+c^2)>=(a+b+c)^@

Mình có cách khác giải thích cho bạn chỗ thắc mắc:

Áp dụng BĐT Cô-si:

   \(a^2+b^2\ge2ab\)

   \(b^2+c^2\ge2bc\)

   \(c^2+a^2\ge2ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)+a^2+b^2+c^2\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)