Đáp án:
$\begin{array}{l}
a)3{x^2} - 5x = 0\\
\Rightarrow x\left( {3x - 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{5}{3}
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{5}{3}\\
b)2{x^2} - 3x - 2 = 0\\
\Rightarrow 2{x^2} - 4x + x - 2 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {2x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = 2;x = - \dfrac{1}{2}\\
c) - 2{x^2} + 8 = 0\\
\Rightarrow {x^2} = 4\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - 2
\end{array} \right.\\
Vậy\,x = 2;x = - 2\\
d){x^4} - 4{x^2} - 5 = 0\\
\Rightarrow \left( {{x^2} - 5} \right)\left( {{x^2} + 1} \right) = 0\\
\Rightarrow {x^2} = 5\\
\Rightarrow x = \pm \sqrt 5 \\
Vậy\,x = \pm \sqrt 5 \\
e){x^4} - 8{x^2} - 48 = 0\\
\Rightarrow \left( {{x^2} - 12} \right)\left( {{x^2} + 4} \right) = 0\\
\Rightarrow {x^2} = 12\\
\Rightarrow x = \pm 2\sqrt 3 \\
Vậy\,x = \pm 2\sqrt 3 \\
f)2{x^4} - 5{x^2} + 2 = 0\\
\Rightarrow \left( {2{x^2} - 1} \right)\left( {{x^2} - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{x^2} = \dfrac{1}{2}\\
{x^2} = 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ \pm \sqrt 2 }}{2}\\
x = \pm \sqrt 2
\end{array} \right.\\
g){x^2} + x - 2 = 0\\
\Rightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = - 2\\
x = 1
\end{array} \right.\\
h)3{x^4} - 12{x^2} + 9 = 0\\
\Rightarrow \left( {3{x^2} - 3} \right)\left( {{x^2} - 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{x^2} = 1\\
{x^2} = 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \pm 1\\
x = \pm \sqrt 3
\end{array} \right.\\
i)16{x^2} + 8x + 1 = 0\\
\Rightarrow {\left( {4x + 1} \right)^2} = 0\\
\Rightarrow x = - \dfrac{1}{4}\\
Vậy\,x = - \dfrac{1}{4}\\
j)Dkxd:x \ne 1;x \ne - 1\\
\dfrac{{12}}{{x - 1}} - \dfrac{8}{{x + 1}} = 1\\
\Rightarrow \dfrac{{12\left( {x + 1} \right) - 8\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 1\\
\Rightarrow 12x + 12 - 8x + 8 = {x^2} - 1\\
\Rightarrow {x^2} - 4x - 21 = 0\\
\Rightarrow \left( {x - 7} \right)\left( {x + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 7\\
x = - 3
\end{array} \right.\left( {tmdk} \right)
\end{array}$
