Question

[Uzumaki Naruto]

Cho \(a^3+b^3+c^3=3abc\left(a,b,c>0\right)\)

Tính \(B=\frac{a^2+2b^2+3c^2}{6ab}\)

Answer 1

[l҉o҉n҉g҉ d҉z҉]

a³ + b³ + c³ = 3abc

<=> a³ + b³ + c³ - 3abc = 0

<=> ( a³ + b³ ) + c³ - 3abc = 0

<=> ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

<=> [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

<=> ( a + b + c )[ ( a + b )² - ( a + b )c + c² ] - 3ab( a + b + c ) = 0

<=> ( a + b + c )( a² + 2ab + b² - ac - bc + c² - 3ab ) = 0

<=> ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

Vì a,b,c > 0 => a + b + c > 0

=> a² + b² + c² - ab - bc - ac = 0

<=> 2( a² + b² + c² - ab - bc - ac ) = 2.0

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( a² - 2ac + c² ) = 0

<=> ( a - b )² + ( b - c )² + ( a - c )² = 0

<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Rightarrow a=b=c\)

<=> \(B=\frac{a^2+2a^2+3a^2}{6a\cdot a}=\frac{6a^2}{6a^2}=1\)