Toán Lớp 8: b) x^3 − 8 + 6x(2 − x) = x − 2

Question

Toán Lớp 8:  b) x^3 − 8 + 6x(2 − x) = x − 2

buithaianh98 · Answer

$ text{b) x&sup3; - 8 + 6x(2 - x) = x - 2}$   $ text{&hArr; (x&sup3; - 8) - 6x(x - 2) - (x - 2) = 0}$   $ text{&hArr; (x - 2)(x&sup2; + 2x + 4) - 6x(x - 2) - (x - 2) = 0}$   $ text{&hArr; (x - 2)(x&sup2; + 2x + 4 - 6x - 1) = 0}$   $ text{&hArr; (x - 2)(x&sup2; - 4x + 3) = 0}$   $ text{&hArr; (x - 2)(x&sup2; - x - 3x + 3) = 0}$   $ text{&hArr; (x - 2)[x(x - 1) - 3(x - 1)] = 0}$   $ text{&hArr; (x - 2)(x - 1)(x - 3) = 0}$   $ text{&hArr; $ left[ begin{matrix} x - 2 = 0  x - 1 = 0  x - 3 = 0 end{matrix} right.$}$   $ text{&hArr; $ left[ begin{matrix} x = 2   x = 1  x = 3 end{matrix} right.$}$   $ text{Vậy x &isin; {1; 2; 3}}$   $ textit{Ha1zzz}$

dongoctuan · Answer

b) x^3 - 8 + 6x ( 2-x) = x-2 <=> x^3 -8 + 12x - 6x^2 = x-2 <=> x^3 -8 + 12x - 6x^2 - x+2=0 <=> x^3- 6x^2 + 11x - 6=0 <=>x^3 - 2x^2- 4x^2 +8x +3x -6=0 <=> x^2(x-2) - 4x(x-2) + 3(x-2)=0 <=> (x-2)(x^2 - 4x+3)=0 <=> (x-2) (x^2 - 3x -x +3)=0 <=> (x-2)[x(x-3) - (x-3)] =0 <=> (x-2)(x-3)(x-1)=0 <=> ( left[ begin{array}{l}x-2=0 x-3=0 x-1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=2 x=3 x=1 end{array} right. ) Vậy x in { 2;3;1}