Đáp án:
2) a) x=2
b) x=-4
c) x=1
Giải thích các bước giải:
Lần sau em tách từng câu ra nhé
\(\begin{array}{l}
a)\,DK:\,{x^2} + {y^2} \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
y \ne 0
\end{array} \right.\\
b)\,DK:{\left( {x - 1} \right)^2} \ne 0 \Leftrightarrow x \ne 1\\
c)\,DK:\,{x^2} + 6x + 10 \ne 0 \Leftrightarrow {\left( {x + 3} \right)^2} + 1 \ne 0\,\left( {ld} \right)\\
\Rightarrow x \in R\\
d)DK:{\left( {x + 3} \right)^2} + {\left( {y - 2} \right)^2} \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne - 3\\
y \ne 2
\end{array} \right.\,\\
2)\,a)\,DK:\,{x^2} + 3x - 10 \ne 0 \Leftrightarrow \left( {x + 2} \right)\left( {x - 5} \right) \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne - 2\\
x \ne 5
\end{array} \right.\\
ycbt \Rightarrow {x^2} - 4 = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x = - 2\left( {ktm} \right)
\end{array} \right.\\
b)\,DK:\,x\left( {x - 4} \right)\left( {x + 1} \right) \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne 4\\
x \ne - 1
\end{array} \right.\\
ycbt \Rightarrow {x^3} - 16x = 0 \Leftrightarrow x\left( {x - 4} \right)\left( {x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\left( L \right)\\
x = - 4\left( N \right)\\
x = 4\left( L \right)
\end{array} \right.\\
c)\,DK:x \ne 1\\
ycbt \Rightarrow {x^3} + {x^2} - x - 1 = 0\\
\Leftrightarrow {x^2}\left( {x + 1} \right) - \left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {{x^2} - 1} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\left( L \right)\\
x = - 1\left( N \right)
\end{array} \right.
\end{array}\)
