a, $x^4-17x^2+16=0$
$⇔x^4-x^2-16x^2+16=0$
$⇔x^2(x^2-1)-16(x^2-1)=0$
$⇔(x^2-16)(x^2-1)=0$
\(⇔\left[ \begin{array}{l}x^2-16=0\\x^2-1=0\end{array} \right.⇔\) \(\left[ \begin{array}{l}x^2=16\\x^2=1\end{array} \right.⇔\) \(\left[ \begin{array}{l}x=±4\\x=±1\end{array} \right.\)
Vậy $x=±4;±1$
b, $4x^4-29x^2+45=0$
$⇔4x^4-20x^2-9x^2+45=0$
$⇔4x^2(x^2-5)-9(x^2-5)=0$
$⇔(4x^2-9)(x^2-5)=0$
\(⇔\left[ \begin{array}{l}4x^2-9=0\\x^2-5=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x^2=\frac{9}4\\x^2=5\end{array} \right.⇔\) \(\left[ \begin{array}{l}x=±\frac{3}2\\x=±\sqrt{5}\end{array} \right.\)
Vậy $x=±\frac{3}2;±\sqrt{5}$
