Đáp án:
$g)
G=\left\{ 1;\dfrac{1}{6} \right \}\\
h)
H=\{-1;0;1;3;7\}\\
i)
I=\{1;2;3\}\\
j)
J=\{0;2;\dfrac{-1}{2}\}$
Giải thích các bước giải:
$g)
G=\{x\in \mathbb{Q}| (6x^2-7x+1)(x^2-5)=0\}\\
(6x^2-7x+1)(x^2-5)=0\\
\Leftrightarrow {\left[\begin{aligned}6x^2-7x+1=0\\x^2-5=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=1\\ x=\dfrac{1}{6}\\x=\sqrt{5}\\ x=-\sqrt{5}\end{aligned}\right.}$
Vì $x\in \mathbb{Q} \Rightarrow G=\left \{1;\dfrac{1}{6} \right \}$
$h)
H=\{x\in \mathbb{R}| (x^2-10x+21)(x^3-x)=0\}\\
(x^2-10x+21)(x^3-x)=0\\
\Leftrightarrow {\left[\begin{aligned}x^2-10x+21=0\\x^3-x=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=7\\ x=3\\x=0\\ x=1\\x=-1\end{aligned}\right.}\\
\Rightarrow H=\{-1;0;1;3;7\}\\
i)
I=\{x\in \mathbb{Z}^+| (6x^2-7x+1)(x^2-5x+6)=0\}\\
(6x^2-7x+1)(x^2-5x+6)=0\\
\Leftrightarrow {\left[\begin{aligned}6x^2-7x+1=0\\x^2-5x+6=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=1\\ x=\dfrac{1}{6}\\x=3\\x=2\end{aligned}\right.}$
Vì $x\in \mathbb{Z}^+\Rightarrow I=\{1;2;3\}$
$j)
J=\{x\in \mathbb{Q}| (2x-x^2)(2x^2-3x-2)=0\}\\
(2x-x^2)(2x^2-3x-2)=0\\
\Leftrightarrow {\left[\begin{aligned}2x-x^2=0\\2x^2-3x-2=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=2\\ x=0\\x=-\dfrac{1}{2}\\x=2\end{aligned}\right.}\\
\Rightarrow J=\left\{0;2;\dfrac{-1}{2}\right \}$
