Question

Tìm x, biết:

a) x³ - 5x² + 8x - 4= 0

b) 2x³ - x² + 3x + 6= 0

c) (x² + x) (x² + x + 1) = 6

d) (x² - 4x)² - 8(x² - 4x) + 15=0

Answer 1

a) \(x^3-5x^2+8x-4=0\)

\(\Leftrightarrow x^3-x^2-4x^2+4x+4x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy nghiệm của phương trình là: \(x=\left\{1;2\right\}\)

Answer 2

b: =>2x^3+2x^2-3x^2-3x+6x+6=0

=>(x+1)(2x^2-3x+6)=0

=>x+1=0

=>x=-1

c: =>(x^2+x)^2+(x^2+x)-6=0

=>(x^2+x-2)=0

=>(x+2)(x-1)=0

=>x=1 hoặc x=-2

d: =>(x^2-4x-3)(x^2-4x-5)=0

=>(x-5)(x+1)(x^2-4x-3)=0

hay \(x\in\left\{2+\sqrt{7};2-\sqrt{7};5;-1\right\}\)