Question

Tìm x : 

a) x^3 - \(3x^2\) + 3x - 1 = 8

b) ( x - 2 )^3 + 6( x + 1 )^2 - x^3 = -12

Answer 1

a)x³-3x²+3x-1=8

=>x³-3x²+3x-9=0

=>x³+3x-3x²-9=0

=>x(x²+3)-3(x²+3)=0

=>(x-3)(x²+3)=0

=>x-3=0 hoặc x²+3=0

Vì x²+3>0 với mọi x =>vô nghiệm

=>x=3 

 

Answer 2

a) \(x^3-3x^2+3x-1=8\)

\(\Leftrightarrow\left(x-1\right)^3=8\)

\(\Leftrightarrow x-1=2\)

\(\Leftrightarrow x=3\)

b) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3=-12\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3=-12\)

\(\Leftrightarrow24x=-10\)

\(\Leftrightarrow x=-\frac{5}{12}\)

Answer 3

b) ( x - 2 )³ + 6( x + 1 )² - x³ = -12

=>x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

=>-6x²+12x+4+6x²+12x+6=0

=>24x+10=0

=>-24x=-10

=>x=\(-\frac{5}{12}\)