Đáp án:

Giải thích các bước giải:

Ta có:

 ` ( 2012a + b + c + d )/( a ) = ( a + 2012b + c + d )/( b ) = ( a + b + 2012c + d )/( c ) = ( a + b + c + 2012d )/( d ) `

` ( 2012a + b + c + d )/( a ) - 2011 = ( a + 2012b + c + d )/( b ) - 2011 = ( a + b + 2012c + d )/( c ) - 2011 = ( a + b + c + 2012d )/( d ) - 2011  `

` ( 2012a + b + c + d )/( a ) - ( 2011a )/( a ) = ( a + 2012b + c + d )/( b ) - ( 2011b )/( b ) = ( a + b + 2012c + d )/( c ) - ( 2011c )/( c ) = ( a + b + c + 2012d )/( d ) - ( 2011d )/( d ) `

` => ( a + b + c+ d )/( a ) = ( a + b + c + d )/( b ) = ( a + b + c + d )/( c ) = ( a + b + c + d )/( d ) `

` TH1: `

` a + b + c + d ne 0 `

` => a = b = c = d `

Thay ` a = b = c = d ` vào ` M ` có:

` M = ( a + b )/( c + d ) + ( b + c )/( d + a ) + ( c + d )/( a + b ) + ( d + a )/( b + c ) `

` M = ( 2a )/( 2a ) + ( 2a )/( 2a ) + ( 2a )/( 2a ) + ( 2a )/( 2a ) `

` M = 1 + 1 + 1 + 1 `

` M = 4 `

` TH2: `

` a + b + c + d = 0 `

` => a + b = -( c + d ) `

` => b + c = -( d + a ) `

` => c + d = -( a + b ) `

` => d + a = -( b + c ) `

Thay vào ` M ` có:

` M = ( -( c + d ) )/( c + d ) + ( -( d + a ) )/( d + a ) + ( -( a + b ) )/( a + b ) + ( -( b + c ) )/( b + c ) `

` M = ( -1 ) + ( -1 ) + ( -1 ) + ( -1 ) `

` M = -4 `

Vậy ` M in { 4; -4 } `

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