Question

a) (x + 3) (x² - 3x + 5) = x² + 3x        Tìm x:

b) x⁴ - 9x² = 4x² - 36

Answer 1

`a) (x+3)(x^2 -3x+5) =x^2 +3x`

`<=> (x+3)(x^2 -3x+5) - x^2 -3x =0`

`<=> (x+3)(x^2 -3x+5) - x(x+3) =0`

`<=>(x+3)(x^2-3x+5-x) =0`

`<=>(x+3)(x^2 -4x+5)=0`

`TH1 : x+3 =0 => x=-3`

`TH2 : x^2 -4x +5 =0 `

`=> Delta = (-4)^2 -4*1*5 =-4 <0`

`=>` PT vô nghiệm

Vậy `S={-3}`

`b) x^4 -9x^2 = 4x^2 -36`

`<=> x^4 -9x^2 - 4x^2 +36 =0`

`<=> x^2(x^2-9) - 4(x^2-9) =0`

`<=> (x^2-4)(x^2-9) =0`

`TH1 : x^2 -4 =0 <=> x^2 =4 <=> x=+-2`

`TH2 : x^2 -9 =0 <=> x^2 =9 <=> x=+-3`

Vậy `S={+-2 ; +-3}`