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         ;( $\forall$ a, b, c )}$

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

   $\text{→ Dấu  ” = ” xảy ra khi :}$

   $\text{$\begin{cases} a-c=0\\b-c=0 \end{cases}$}$

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

   $\text{⇒ a = b = c}$

   $\text{→ Vậy a = b = c            ( ĐPCM )}$

   5 sao nha

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