Bài `38:`
`a) (x-2)^3 - x . (x+1).(x-1) + 6x . (x-3)`
` = x^3 -3.x^2.2 + 3 . x . 2^2 - 2^3 - x . (x^2-1) + 6x^2 - 18x`
` = x^3 - 6x^2 + 12x - 8 - x^3 + x + 6x^2 - 18x`
` = (x^3 - x^3) + (6x^2 - 6x^2) + (12x - 18x+x) - 8`
` = -5x - 8`
`b) (x-2).(x^2- 2x+4) . (x+2).(x^2 + 2x+4)`
` = [ (x-2). (x+2)] . [ (x^2 - 2x + 4).(x^2 + 2x+4)]`
` = (x^2 - 4). (x^4 + 4x^2 + 16)`
` = (x^2)^3 - 4^3`
` = x^6 - 64`
Bài `39:`
`a) (x-3).(x^2 + 3x+9) + x . (x+2).(2-x) = 1`
`=> x^3 - 3^3 + x . (4 - x^2) =1`
`=> x^3 - 27 + 4x - x^3 = 1`
=> 4x - 27 =1`
`=> 4x = 28`
`=> x = 7`
Vậy `x=7`
`b) (x+1)^3 - (x-1)^3 - 6(x-1)^2 = -10`
`=>( x^3 + 3x^2 + 3x + 1)-(x^3 - 3x^2 + 3x-1) - 6 . (x^2 - 2x + 1) = -10`
`=> x^3 + 3x^2+ 3x+1 - x^3 + 3x^2 - 3x + 1 - 6x^2 + 12x - 6 = -10`
`=> (x^3 - x^3) + (3x^2 + 3x^2 - 6x^2) + (3x-3x+12x) + (1+1-6) = -10`
`=> 12x - 4 = -10`
`=> 12x = -6`
`=> x = -1/2`
Vậy `x=-1/2`
Bài `40:`
`a)`
Đặt `A = (a+b+c)^3 - (b+c-a)^3 - (a+c-b)^3 - (a+b-c)^3`
Đặt `b+c-a = x ; a+c-b = y ; a+b-c=z `
Khi đó ta có :
`x + y + z =b+c-a+a+c-b+a+b-c = a+b+c`
Do đó ta có :
`A = (x+y+z)^3 - x^3 - y^3 - z^3`
` = 3 . (x+y).(y+z).(x+z)`
Mà `b+c-a = x ; a+c-b = y ; a+b-c=z ` nên ta có :
`A = 3 . (b+c-a+a+c-b) . (a+c-b + a+b-c) . (b+c-a+a+b-c)`
` = 3 . 2c . 2a . 2b`
` = 24abc`
`b) (a+b)^3 + (b+c)^3 + (c+a)^3 - 3 . (a+b).(b+c).(c+a)`
`= (a+b+b+c+a+c) . [ (a+b)^2 + (b+c)^2 + (a+c)^2 - (a+b)(b+c) - (b+c)(c+a) - (c+a)(a+b) ]`
` = (2a + 2b + 2c) . [ (a^2 + 2ab + b^2) + (b^2 + 2bc + c^2) + (a^2 + 2ac + c^2) - (ab + ac + b^2 + bc) - (bc + ab + c^2 + ac) - (ac + bc + a^2 + ab)]`
` = 2 . (a+b+c) . (a^2 + b^2 + c^2 - ab- ac-bc)`
