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Bài `6 :`

`a) A = 1/( y^2 - xy ) + 1/( x^2 - xy )`

`= 1/( y ( y - x ) ) + 1/( x ( x - y ) )`

`= 1/( y ( y - x ) ) - 1/( x ( y - x ) )`

`= x/( xy ( y - x ) ) - y/( xy ( y - x ) )`

`= ( x - y )/( xy ( y - x ) )`

`= ( x - y )/( - 1 . ( y - x ) )`

`= ( x - y )/( x - y )`

`= 1`

`b) B = x^3/( ( x - y ) ( x - z ) ) + y^3/( ( y - z )( y - x ) ) + z^3/( ( z - x ) ( z - y ) )`

`= ( - x^3 ( y - z ) )/( ( x - y ) ( y - z ) ( z - x ) ) - ( y^3 ( z - x ) )/( ( x - y ) ( y - z ) ( z - x ) ) - ( z^3  ( x - y ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= ( -x^3 ( y - z ) - y^3 ( z - x ) - z^3 ( x - y ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= - ( x^3 ( y - z ) + y^3 ( z - x ) + z^3 ( x - y ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= - ( x^3 ( y - z ) + y^3 ( z - x ) + z^3 ( x - z + z - y ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= - ( x^3 ( y - z ) + y^3 ( z - x ) +  z^3 ( x - z ) + z^3 ( z - y ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= - ( ( y - z ) ( x^3 - z^3 ) + ( z - x ) ( y^3 - z^3 ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= - ( ( y - z ) ( x - z ) ( x^2 + xz + z^2 ) + ( z - x ) ( y - z ) ( y^2 + yz + z^2 ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= - ( ( y - z ) ( x - z ) ( x^2 + xz + z^2 ) - ( x - z ) ( y - z ) ( y^2 + yz + z^2 ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= - ( ( y - z ) ( x - z ) ( x^2 + xz + z^2 - y^2 - yz - z^2 ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= - ( ( y - z ) ( x - z ) ( ( x - y ) ( x + y ) + z ( x - y ) ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= - ( ( y - z ) ( x - z ) ( x - y ) ( x + y + z ) )/( ( x - y ) ( y - z ) ( z - x ) )`

`= x + y + z = 2017`

`c)` Ta có `: ( 2018a )/( ab + 2018a + 2018 ) = ( a^2bc )/( ab + a^2bc + abc ) = ( ac )/( 1 + ac + c )`

                 `b/( bc  +b + 2018 ) = b/( bc + b  +abc ) = 1/( c + 1 + ac )`

`=> C = ( ac )/( ac + c + 1 ) + 1/( c + 1 + ac ) + c/( ac + c + 1 ) = ( ac + c + 1 )/( ac + c  +1 ) = 1`

Đáp án:

Giải thích các bước giải:

 Bài `6:`
`a)A=1/(y^2−xy)+1/(x^2−xy)`
`=1/(y(y−x))+1/(x(x−y)`
`=1/(y(y−x)−1/(x(y−x))`
`=x/(xy(y−x)) – y/(xy(y−x)` 
`=(x – y)/( x y ( y − x ))`
` =( x – y)/(1 ( x-y) )`
`= 1`
`b)B=x^3/((x−y)(x−z))+y^3/((y−z)(y−x))+z^3/((z−x)(z−y))`
`=(-x^3(y−z))/((x−y)(y−z)(z−x))–(y^3(z−x))/((x−y)(y−z)(z−x))–(z^3(x−y))/((x−y)(y−z)(z−x))`
`=(-x^3(y−z)−y^3(z−x)−z^3(x−y))/((x−y)(y−z)(z−x))`
`=(-x^3(y−z)+y^3(z−x)+z^3(x−y))/((x−y)(y−z)(z−x))`
`=(−x^3(y−z)+y^3(z−x)+z^3(x−z+z−y))/((x−y)(y−z)(z−x))`
`=(-x^3(y−z)+y^3(z−x)+z^3(x−z)+z^3(z−y))/((x−y)(y−z)(z−x))`
`=(−(y−z)(x^3−z^3)+(z−x)(y^3−z^3))/((x−y)(y−z)(z−x))`
`=(-(y−z)(x−z)(x^2+xz+z^2)−(x−z)(y−z)(y^2+yz+z^2))/((x−y)(y−z)(z−x))`
`=(−(y−z)(x−z)(x^2+xz+z^2−y^2−yz−z^2))/((x−y)(y−z)(z−x))`
`=(−(y−z)(x−z)[(x−y)(x+y)+z(x−y)])/((x−y)(y−z)(z−x))`
`=(−(y−z)(x−z)(x−y)(x+y+z))/((x−y)(y−z)(z−x)`
`=x+y+z`
`=2017`
 `c)` Ta có :
 `(2018a)/(ab+2018a+2018=(a^2bc)/(ab+a^2bc+abc)=(ac)/(1+ac+cb)`
`b/(c+b+2018)=b
/(bc+b+abc)=1/(c+1+ac)`
Từ đó suy ra:
`C=(ac)/(ac+c+1)+1/(c+1+ac)+c/(ac+c+1)=(ac+c+1)/(ac+c+1)=1`