Cho `(a – b)^2 + ( b-c)^2 + ( c-a)^2 = ( a + b – 2c)^2 + ( b + c – 2a)^2 + ( c +a – 2b )^2` CMR : ` a= b = c`

1 bình luận về “Cho `(a – b)^2 + ( b-c)^2 + ( c-a)^2 = ( a + b – 2c)^2 + ( b + c – 2a)^2 + ( c +a – 2b )^2` CMR : ` a= b = c`”

1. $\text{→ Ta có :}$

   $\text{( a – b )² + ( b – c )² + ( c – a )² = ( a + b – 2c )² + ( b + c – 2a )² + ( c + a – 2b )²}$

   $\text{⇔ ( a – b )² – ( a + b – 2c )² + ( b – c )² – ( b + c – 2a )² + ( c – a )² – ( c + a – 2b )² = 0}$

   $\text{⇔ ( a – b + a + b – 2c )( a – b – a – b + 2c ) + ( b – c + b + c – 2a )( b – c – b – c + 2a ) +}$

   $\text{( c – a + c + a – 2b )( c – a – c – a + 2b ) =0}$

   $\text{⇔ ( 2a – 2c )( 2c – 2b ) + ( 2b – 2a )( 2a – 2c ) + ( 2c – 2b )( 2b – 2c ) =0}$

   $\text{⇔ 4( a – c )( c – b ) + 4( b – a )( a – c ) + 4( c – b )( b – c ) =0}$

   $\text{⇔ ( a – c )( c – b ) + ( b – a )( a – c ) + ( c – b )( b – c ) =0}$

   $\text{⇔ ( a – c )( c – b + b – a ) – ( b – c )( b – c ) = 0}$

   $\text{⇔ ( a – c )( c – a ) – ( b – c )² = 0}$

   $\text{⇔ – ( a – c )( a – c ) – ( b – c )² =0}$

   $\text{⇔ – ( a – c )( a – c ) – ( b – c )² =0}$

   $\text{⇔ ( a – c )² + ( b – c )² = 0}$

   $\text{mà ( a – c )² ≥ 0 và ( b – c )² ≥ 0 ;( }\mathrm{\forall }\text{ a, b, c )}$

   $\text{⇒ ( a – c )² + ( b – c )² ≥ 0 ;( }\mathrm{\forall }\text{ a, b, c )}$

   $\text{→ Dấu ” = ” xảy ra khi :}$

   $\{\begin{array}{l}a-c=0\\ b-c=0\end{array}$

   $\text{⇔ }\{\begin{array}{l}a=c\\ b=c\end{array}$

   $\text{⇒ a = b = c}$

   $\text{→ Vậy a = b = c ( ĐPCM )}$

   5 sao nha

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