Đáp án:
a) $x\in \left \{ 0;5 \right \}$
b) $x\in \left \{ -6;7 \right \}$
c) $x\in \left \{ -2;8 \right \}$
d) $x\in \varnothing $
Giải thích các bước giải:
a) $3x^{2}-15x=0$
$\leftrightarrow 3x(x-5)=0$
$\leftrightarrow x(x-5)=0$
$\leftrightarrow \left[ \begin{array}{l}x=0\\x-5=0\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}x=0\\x=5\end{array} \right.$
Vậy $x\in \left \{ 0;5 \right \}$
b) $x(x+6)-7x-42=0$
$\leftrightarrow x(x+6)-7(x+6)=0$
$\leftrightarrow (x+6)(x-7)=0$
$\leftrightarrow \left[ \begin{array}{l}x+6=0\\x-7=0\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}x=-6\\x=7\end{array} \right.$
Vậy $x\in \left \{ -6;7 \right \}$
c) $25-(3-x)^{2}=0$
$\leftrightarrow 5^{2}-(3-x)^{2}=0$
$\leftrightarrow (5-3+x)(5+3-x)=0$
$\leftrightarrow (x+2)(8-x)=0$
$\leftrightarrow \left[ \begin{array}{l}x+2=0\\8-x=0\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}x=-2\\x=8\end{array} \right.$
Vậy $x\in \left \{ -2;8 \right \}$
d) $4x^{2}+11x+9=0$
$\leftrightarrow (2x)^{2}+12x+9-x=0$
$\leftrightarrow (2x+9)^{2}-x=0$
$\leftrightarrow (2x+9-\sqrt{x})(2x+9+\sqrt{x})=0$
$\leftrightarrow \left[ \begin{array}{l}2x+9-\sqrt{x}=0\\2x+9+\sqrt{x}=0\end{array} \right.$
$\leftrightarrow \left[ \begin{array}{l}x\in \varnothing \\x\in \varnothing \end{array} \right.$
Vậy $x\in \varnothing $
