Đáp án:

 $a) x=2\\
b)  {\left[\begin{aligned}x=0\\x=6\end{aligned}\right.}\\
c) {\left[\begin{aligned}x=3\\x=4\end{aligned}\right.}\\
d){\left[\begin{aligned}x=0\\x=4\\ x=-4\end{aligned}\right.}\\
e)w {\left[\begin{aligned}x=8\\x=1\end{aligned}\right.}\\
f)  {\left[\begin{aligned}x=-1\\x=\frac{-4}{5}\end{aligned}\right.}\\$

Giải thích các bước giải:

 $a)f(x)=3x-6=0\\
\Leftrightarrow 3(x-2)=0\Leftrightarrow x=2\\
b) g(x)=-5x^2+30x=0\\
\Leftrightarrow -5x(x-6)=0\\
\Leftrightarrow {\left[\begin{aligned}x=0\\x=6\end{aligned}\right.}\\
c)h(x)=(x-3)(16-4x)=0\\
\Leftrightarrow {\left[\begin{aligned}x-3=0\\16-4x=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=3\\x=4\end{aligned}\right.}\\
d)k(x)=2x^3-32x=0\\
\Leftrightarrow 2x(x^2-16)=0\\
\Leftrightarrow {\left[\begin{aligned}x=0\\x^2=16\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=0\\x=4\\ x=-4\end{aligned}\right.}\\
e)m(x)=x^2+7x-8=0\\
\Leftrightarrow x^2+8x-x-8=0\\
\Leftrightarrow x(x+8)-(x-8)=0\\
\Leftrightarrow (x-8)(x-1)=0\\
\Leftrightarrow {\left[\begin{aligned}x=8\\x=1\end{aligned}\right.}\\
f) n(x)=5x^2+9x+4=0\\
\Leftrightarrow 5x^2+5x+4x+4=0\\
\Leftrightarrow 5x(x+1)+4(x+1)=0\\
\Leftrightarrow (x+1)(5x+4)=0\\
\Leftrightarrow {\left[\begin{aligned}x+1=0\\5x+4=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=-1\\x=\frac{-4}{5}\end{aligned}\right.}$

$f(x)=3x-6=0$

⇒ $3x=6$

⇒ $x=2$

$g(x)=-5x^2+30x=0$

⇒\(\left[ \begin{array}{l}x=0\\-5x+30=0\end{array} \right.\) $x.(-5x+30)=0$

⇒\(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\)

$h(x)=(x-3)(16-4x)=0$

⇒\(\left[ \begin{array}{l}x-3=0\\16-3x=0\end{array} \right.\)

⇒\(\left[ \begin{array}{l}x=3\\x=\dfrac{16}{3}\end{array} \right.\)

$k(x)=2x^3-32x$

⇒\(\left[ \begin{array}{l}x=0\\2x^2-3=0\end{array} \right.\)

⇒\(\left[ \begin{array}{l}x=0\\x=\dfrac{\sqrt{6}}{2}or\dfrac{-\sqrt{6}}{2}\end{array} \right.\)

$m(x)=x^2+7x-8$

⇒ $x^2+8x-x-8$

⇒ $x(x-1)+8.(x-1)$

⇒ $(x-1).(x+8)=0$

⇒\(\left[ \begin{array}{l}x-1=0\\x+8=0\end{array} \right.\)

⇒\(\left[ \begin{array}{l}x=1\\x=-8\end{array} \right.\)

$n(x)=5x^2+9x+4$

⇒ $(x+1).(5x+4)=0$

⇒\(\left[ \begin{array}{l}x+1=0\\5x+4=0\end{array} \right.\)

⇒\(\left[ \begin{array}{l}x=-1\\x=\dfrac{-4}{5}\end{array} \right.\)