Question

[Hà Nam Khánh]

Giải các phương trình và bất phương trình sau:

1) x⁴ + 3x² - 4 = 0

2) (x² - 2x)² + |x² - 2x| - 2 = 0

3) \(\dfrac{x}{x+2}\) < \(\dfrac{x}{x+1}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

1: \(x^4+3x^2-4=0\)

=>\(x^4+4x^2-x^2-4=0\)

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

=>\(x^2-1=0\)

=>\(x^2=1\)

=>\(x=\pm1\)

2: \(\left(x^2-2x\right)^2+\left|x^2-2x\right|-2=0\)

=>\(\left(\left|x^2-2x\right|\right)^2+\left|x^2-2x\right|-2=0\)

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

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

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

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

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

=>\(x\in\left\{1;\pm\sqrt{2}+1\right\}\)

3: ĐKXĐ: \(x\notin\left\{-2;-1\right\}\)

\(\dfrac{x}{x+2}< \dfrac{x}{x+1}\)

=>\(\dfrac{x}{x+2}-\dfrac{x}{x+1}< 0\)

=>\(\dfrac{x\left(x+1\right)-x\left(x+2\right)}{\left(x+2\right)\left(x+1\right)}< 0\)

=>\(\dfrac{-x}{\left(x+2\right)\left(x+1\right)}< 0\)

=>\(\dfrac{x}{\left(x+1\right)\left(x+2\right)}>0\)

TH1: \(\left\{{}\begin{matrix}x>0\\\left(x+1\right)\left(x+2\right)>0\end{matrix}\right.\)

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

=>\(x>0\)

TH2: \(\left\{{}\begin{matrix}x< 0\\\left(x+1\right)\left(x+2\right)< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 0\\-2< x< -1\end{matrix}\right.\)

=>-2<x<-1

Answer 2

[Akai Haruma]

1.

$x^4+3x^2-4=0$

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

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

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

$\Leftrightarrow x^2-1=0$ hoặc $x^2+4=0$

Nếu $x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm 1$

Nếu $x^2+4=0\Leftrightarrow x^2=-4<0$ (vô lý)

Vậy pt có nghiệm $x=1$ hoặc $x=-1$