Đáp án:
$a)
A=\left \{ \dfrac{1}{3};\dfrac{-1}{2} \right \}\\
b)
B=\{-2;-1;1;2\}\\
c)
C=\{-\sqrt{3};\sqrt{3};2\}\\
d)
D=\{1\}\\
e)
E=\{-1;2\}\\
f)
F=\{-1;3\}\\
g)
F=\{1;2;\dfrac{-3+\sqrt{17}}{2};\dfrac{-3-\sqrt{17}}{2}\}\\$
Giải thích các bước giải:
$a)
A=\{x\in \mathbb{R}/ 6x^2+x-1=0\}\\
6x^2+x-1=0\Leftrightarrow {\left[\begin{aligned}x=\dfrac{1}{3}\\x=\dfrac{-1}{2}\end{aligned}\right.}\\
\Rightarrow A=\left \{ \dfrac{1}{3};\dfrac{-1}{2} \right \}\\
b)
B=\{x\in \mathbb{R}/ x^4-5x^2+4=0\}\\
x^4-5x^2+4=0 \Leftrightarrow {\left[\begin{aligned}x=2\\x=1\\ x=-1\\ x=-2\end{aligned}\right.}\\
\Rightarrow B=\{-2;-1;1;2\}\\
c)
C=\{x\in \mathbb{R}/ x^3-2x^2-3x+6=0\}\\
x^3-2x^2-3x+6=0
\Leftrightarrow x^2(x-2)-3(x-2)=0\\
\Leftrightarrow (x-2)(x^2-3)=0
\Leftrightarrow {\left[\begin{aligned}x-2=0\\x^2-3=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=2\\x=\pm \sqrt{3}\end{aligned}\right.}\\
\Rightarrow C=\{-\sqrt{3};\sqrt{3};2\}\\
d)
D=\{x\in \mathbb{R}/ x^3+x^2-2=0\}\\
x^3+x^2-2=0\Leftrightarrow x^3-x^2+2x^2-2x+2x-2=0\\
\Leftrightarrow x^2(x-1)+2x(x-1)+2(x-1)=0\\
\Leftrightarrow (x-1)(x^2+2x+2)=0\\
\Leftrightarrow x=1\\
\Rightarrow D=\{1\}\\
e)
E=\{x\in \mathbb{R}/ x^4-x^2-4x-4=0\}\\
x^4-x^2-4x-4=0\Leftrightarrow {\left[\begin{aligned}x=2\\ x=-1\end{aligned}\right.}\\
\Rightarrow E=\{-1;2\}\\
f)
F=\{x\in \mathbb{R}/ x^4-4x^2-12x-9=0\}\\
x^4-4x^2-12x-9=0\Leftrightarrow {\left[\begin{aligned}x=3\\ x=-1\end{aligned}\right.}\\
\Rightarrow F=\{-1;3\}\\
g)
G=\{x\in \mathbb{R}/x^4-9x^2+12x-4=0\}\\
x^4-9x^2+12x-4=0\Leftrightarrow {\left[\begin{aligned}x=2\\ x=1\\ x=\dfrac{-3+\sqrt{17}}{2}\\ x=\dfrac{-3-\sqrt{17}}{2}\end{aligned}\right.}\\
\Rightarrow F=\{1;2;\dfrac{-3+\sqrt{17}}{2};\dfrac{-3-\sqrt{17}}{2}\}\\$
