tính: a^3 + b^3 + c^3 + d^3 – 3.( ab – cd ). ( c + d) với a + b + c + d= 0 26/07/2021 Bởi aihong tính: a^3 + b^3 + c^3 + d^3 – 3.( ab – cd ). ( c + d) với a + b + c + d= 0
`a+b+c+d = 0` `=> a+b = – (c+d)` `=> (a+b)^3 = -(c+d)^3` `a^3 + 3a^2b + 3ab^2 + b^3 = -( c^3 + 3c^2d + 3cd^2 + d^3)` `a^3 + b^3 + c^3 + d^3 = – 3ab(a+b) – 3cd ( c+d)` `=> a^3 + b^3 + c^3 + d^3 = 3ab(c+d) – 3cd(c+d) (Thay a+b = -(c+d)` `a^3 + b^3 + c^3 + d^3 = 3(c+d)(ab-cd) (đpcm)` Bình luận
`a+b+c+d = 0`
`=> a+b = – (c+d)`
`=> (a+b)^3 = -(c+d)^3`
`a^3 + 3a^2b + 3ab^2 + b^3 = -( c^3 + 3c^2d + 3cd^2 + d^3)`
`a^3 + b^3 + c^3 + d^3 = – 3ab(a+b) – 3cd ( c+d)`
`=> a^3 + b^3 + c^3 + d^3 = 3ab(c+d) – 3cd(c+d) (Thay a+b = -(c+d)`
`a^3 + b^3 + c^3 + d^3 = 3(c+d)(ab-cd) (đpcm)`