Toán Lớp 8: Tìm x, biết:
a) x³ + 3x² + 3x = 1
b) (x -7) (x² – 9x + 20) (x – 2) = 72
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Toán Lớp 8: Tìm x, biết:
a) x³ + 3x² + 3x = 1
b) (x -7) (x² – 9x + 20) (x – 2) = 72
Comments ( 2 )
x³-6x²-6x²+12x+36x-72 =0
(x³-6x²) + (-6x²+36x) + (12x-72) =0
x²(x-6) – 6x.(x-6) + 12(x-6) =0
⇒ $\left \{ {{x²-6x+12=0} \atop {x-6=0}} \right.$
⇒ $\left \{ {{(x+3)² + 3=0} (Vô lí) \atop {x=6}} \right.$
x^3-6x^2-6x^2+12x+36x-72=0
(x^3-6x^2)+(-6x^2+36x)+(12x-72)=0
x^2(x-6)-6x(x-6)+12(x-6)=0
(x^2-6x+12)(x-6)=0
=>x^2-6x+12=0 hoặc x-6=0
Có x^2 -6x+12=(x-3)^2 +3 lớn hơn hoặc bằng 3 với mọi x
=>x-6=0
=>x=6