Question

Giải các phương trình sau:

a) x³+4x2+x-6=0

b) x³-3x²+4=0

c)x⁴+2x³+2x²-2x-3=0

d)x⁴+4x³+4x²-4x-5=0

e)(x-\(\sqrt{2}\) )+3(x-2)=0

f)x-5=(2x-\(\sqrt{5}\) )(x+ \(\sqrt{5}\))

Answer 1

a) x³+4x²+x-6=0

<=> x³+x²-2x+3x²+3x-6=0

<=>x(x²+x-2)+3(x²+x-2)=0

<=>(x+3)(x²+x-2)=0

<=>(x+3)(x²+2x-x-2)=0

<=>(x+3)[x(x+2)-(x+2)]=0

<=>(x+3)(x-1)(x+2)=0

=> x+3=0 hay

x-1=0 hay

x+2=0

<=> x=-3 hay x=1 hay x=-2

Answer 2

b)x³-3x²+4=0

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

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

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

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

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

Answer 3

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

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

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

\(\Leftrightarrow\left(x^2-1\right)\left(x^2+2x+3\right)=0\)

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

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

\(\Rightarrow\left\{\begin{matrix}x-1=0\\\left(x+1\right)^2+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=1\\\left(x+1\right)^2=-2\\x=-1\end{matrix}\right.\)(vô lí, vì bình phương một số không thể là số âm)

Vậy phương trình trên có nghiệm là x=1 và x=-1