Đáp án:
B113:
b) \(x \ne \left\{ { - 8;0;1} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
B112:\\
a){x^2} + 5x + 4 = 0\\
\to {x^2} + x + 4x + 4 = 0\\
\to x\left( {x + 1} \right) + 4\left( {x + 1} \right) = 0\\
\to \left( {x + 1} \right)\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 1\\
x = - 4
\end{array} \right.\\
b){x^2} + 2x + 4 = 0\left( {vô lý} \right)\\
Do:{x^2} + 2x + 4 = {x^2} + 2x + 1 + 3\\
= {\left( {x + 1} \right)^2} + 3 > 0\forall x\\
c){\left( {x + 2} \right)^2} + {\left( {2 - 2y} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 2 = 0\\
2 - 2y = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - 2\\
y = 1
\end{array} \right.\\
B113:\\
a)DK:{x^2} - 2x - 8 \ne 0\\
\to \left( {x - 4} \right)\left( {x + 2} \right) \ne 0\\
\to x \ne \left\{ { - 2;4} \right\}\\
b)DK:{x^3} + 7{x^2} - 8x \ne 0\\
\to x\left( {{x^2} - x + 8x - 8} \right) \ne 0\\
\to x\left( {x + 8} \right)\left( {x - 1} \right) \ne 0\\
\to x \ne \left\{ { - 8;0;1} \right\}\\
c)DK:2x - 1 \ne 0 \to x \ne \dfrac{1}{2}
\end{array}\)
