Đáp án:
`a)`
` x(4-x) - (3+x)(3-x) = 9 +2x`
` => 4x - x^2 - (9-x^2) = 9 +2x`
` => 4x - x^2 - 9 + x^2 = 9 +2x`
` => 4x - 9 - 9 - 2x = 0`
`=> 2x -18 = 0`
`=> 2x = 18 => x = 9`
`b)`
` x^3 -25x = 0`
` => x(x^2 -25) = 0`
TH1
` x = 0`
TH2
` x^2 -25 = 0`
` => x^2 = 25 => x = ± 5`
`c)`
` 6x^2 + 11x + 5 = 0`
` => (6x+5)(x+1) =0`
TH1
` 6x +5 = 0 => 6x = -5 => x= -5/6`
TH2
` x +1 = 0 => x =-1`
Vậy ` x ∈ {-5/6 ; -1}`
