*Lời giải :
Ta có :
`f (x) = -3x^3 + 5/4x + 3x^3 - 1/4x - 1 + x`
`-> f (x) = (-3x^3 + 3x^3) + (5/4x - 1/4x + x) - 1`
`-> f (x) = 2x - 1`
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`f (x) = g (x)`
`⇔ 2x - 1 = -x^2 - 5x + 7`
`⇔ 2x - 1 + x^2 + 5x - 7 = 0`
`⇔ (2x + 5x) + (-1 - 7) + x^2 = 0`
`⇔ 7x - 8 + x^2 = 0`
`⇔ x^2 + 7x - 8 = 0`
`⇔ x^2 + 8x - x - 8 = 0`
`⇔ (x^2 + 8x) - (x + 8) = 0`
`⇔ x (x + 8) - (x + 8) = 0`
`⇔ (x + 8) (x - 1) = 0`
`⇔ x + 8 = 0` hoặc `x - 1 = 0`
`⇔ x = -8` hoặc `x = 1`
Vậy `x=-8,x=1` để `f (x) = g (x)`
