Đáp án:
Giải thích các bước giải:
$1: (4x-10)(24+5x)=0$
$\Leftrightarrow \left\{\begin{matrix}
4x-10=0\\
24+5x=0
\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}
x=\frac{5}{2}\\
x=-\frac{24}{5}
\end{matrix}\right.$
Vậy $x\in \left \{ \frac{5}{2};-\frac{24}{5} \right \}$
$2: 5x(x-1)+3(x-1)=0$
$\Leftrightarrow (x-1)(5x+3)=0$
$\Leftrightarrow \left\{\begin{matrix}
x-1=0\\
5x+3=0
\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}
x=-\frac{3}{5}\\
x=1
\end{matrix}\right.$
Vậy $x\in \left \{ -\frac{3}{5};1 \right \}$
$3: (2x-5)(x+2)(3x-7)=0$
$\Leftrightarrow \left\{\begin{matrix}
2x-5=0\\
x+2=0\\
3x-7=0
\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}
x=\frac{5}{2}\\
x=-2\\
x=\frac{7}{3}
\end{matrix}\right.$
Vậy $x\in \left \{ \frac{5}{2};-2;\frac{7}{3} \right \}$
$4: (2x+1)(3x-2)=(5x-8)(2x+1)$
$\Leftrightarrow 6x^{2}-4x+3x-2=10x^{2}+5x-16x-8$
$\Leftrightarrow 6x^{2}-x-2=10x^{2}-11x-8$
$\Leftrightarrow 6x^{2}-x-2-10x^{2}=11x+8=0$
$\Leftrightarrow -4x^{2}+10x+6=0$
$\Leftrightarrow 2x^{2}-5x-3=0$
$\Leftrightarrow 2x^{2}+x-6x-3=0$
$\Leftrightarrow x(2x+1)-3(2x+1)=0$
$\Leftrightarrow (2x+1)(x-3)=0$
$\Leftrightarrow \left\{\begin{matrix}
2x+1=0\\
x-3=0
\end{matrix}\right. $
$\Leftrightarrow \left\{\begin{matrix}
x=-\frac{1}{2}\\
x=3
\end{matrix}\right.$
Vậy $x\in \left \{ -\frac{1}{2};3 \right \}$
