Đáp án + Giải thích các bước giải:
a) $(2x+1)(x^2+2)=0$
⇔$\(\left[ \begin{array}{l}2x+1=0\\x^2+2=0\end{array} \right.\)$
⇔$\(\left[ \begin{array}{l}x=\frac{-1}{2}\\x=±\sqrt{2}\end{array} \right.\)$
Vậy S=${\frac{-1}{2};±\sqrt{2}}$
c) $(5x-3)^2-(4x-7)^2=0$
$⇔(5x-3-4x+7)(5x-3+4x-7)=0$
$⇔(x+4)(9x-10)=0
$⇔\(\left[ \begin{array}{l}x+4=0\\9x-10=0\end{array} \right.\)$
$⇔\(\left[ \begin{array}{l}x=-4\\x=\frac{10}{9}\end{array} \right.\)$
Vậy S=${-4;\frac{10}{9}}$
b) $(x^2+4)(7x-3)=0$
$⇔\(\left[ \begin{array}{l}x^2+4=0\\7x-3=0\end{array} \right.\)$
$⇔\(\left[ \begin{array}{l}x=±2\\x=\frac{3}{7}\end{array} \right.\)$
Vậy S=${±2;\frac{3}{7}}$
d) $(x+2)^2=9(x^2-4x+4)$
$⇔(x+2)^2=9(x-2)^2$
$⇔(x+2)^2-(9x-18)^2=0$
$⇔(x+2-9x+18)(x+2+9x-18)=0$
$⇔(-8x+20)(10x-16)=0$
$⇔\(\left[ \begin{array}{l}-8x+20=0\\10x-16=0\end{array} \right.\)$
$⇔\(\left[ \begin{array}{l}x=\frac{5}{2}\\x=\frac{8}{5}\end{array} \right.\)$
Vậy S=${\frac{5}{2};\frac{8}{5}}$
Mik xin hay nhất ạ
