Đáp án + Giải thích các bước giải:
$a)\; x^2-9x+20=0$
$⇔ x^2-4x-5x+20=0$
$⇔ (x^2-4x)-(5x-20)=0$
$⇔ x(x-4)-5(x-4)=0$
$⇔ (x-5)(x-4)=0$
\(⇔ \left[ \begin{array}{l}x-5=0\\x-4=0\end{array} \right.\)
\(⇔ \left[ \begin{array}{l}x=5\\x=4\end{array} \right.\)
$b)\; (x-2)(x+3)-3(4x-2)=(x-4)^2$
$⇔x^2+3x-2x-6-12x+6=x^2-8x+16$
$⇔x^2-11x-x^2+8x-16=0$
$⇔-3x-16=0$
$⇔-3x=16$
$⇔x=\dfrac{-16}{3}$
$c)\: |2x-3|-4x-9=0$
$⇔|2x-3|=4x+9$
\(⇔\left[ \begin{array}{l}2x-3=4x+9\\2x-3=-4x-9\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-6\\x=-1\end{array} \right.\)
$d)\; x^3+5x^2+3x-9=0$
$⇔x^3+6x^2+9x-x^2-6x-9=0$
$⇔x\left(x^2+6x+9\right)-\left(x^2+6x+9\right)=0$
$⇔\left(x-1\right)\left(x^2+6x+9\right)=0$
$⇔\left(x-1\right)\left(x+3\right)^2$
\(⇔\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
$e)\; (x^2-2)^2=4x^2+1$
$⇔x^4-4x^2+4=4x^2+1$
$⇔x^4-8x^2+3=0$
Đặt $t=x^2$ ta có:
$t^2-8t+3=0$
$⇔ t^2-8t+16=13$
$⇔ (t-4)^2=13$
$⇔ t-4=\sqrt{13}$
$⇔ t=\sqrt{13}+4$
$⇔ x=\sqrt{\sqrt{13}+4}$
$f)\; x^2-x-20=0$
$⇔x^2+4x-5x-20=0$
$⇔(x^2+4x)-(5x+20)=0$
$⇔x(x+4)-5(x+4)=0$
$⇔(x-5)(x+4)=0$
\(⇔ \left[ \begin{array}{l}x-5=0\\x+4=0\end{array} \right.\)
\(⇔ \left[ \begin{array}{l}x=5\\x=-4\end{array} \right.\)
