2 câu trả lời
${\sqrt[]{4-x²}-x+2=0}$ ${ĐKXĐ:x \geq 2 }$
${⇔\sqrt[]{4-x²}=x-2}$
${⇔4-x²=(x-2)²}$
${⇔4-x²=x²-4x+4}$
${⇔4-x²-x²+4x-4=0}$
${⇔-2x²+4x=0}$
${⇔-2x(x-2)=0}$
${⇔\left[\begin{matrix} -2x=0\\ x-2=0\end{matrix}\right.}$
${⇔\left[\begin{matrix} x=0,ktm\\ x=2,tm\end{matrix}\right.}$
Vậy ${x=2}$
#andy
\[\begin{array}{l}
\sqrt {4 - {x^2}} - x + 2 = 0\\
DKXD:x \ge 2\\
\Leftrightarrow \sqrt {4 - {x^2}} = x - 2\\
\Leftrightarrow 4 - {x^2} = {(x - 2)^2}\\
\Leftrightarrow 4 - {x^2} = {x^2} - 4x + 4\\
\Leftrightarrow 4 - {x^2} - {x^2} + 4x - 4 = 0\\
\Leftrightarrow - 2{x^2} + 4x = 0\\
\Leftrightarrow - 2x(x - 2) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
- 2x = 0\\
x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\left( {KTMDK} \right)\\
x = 2\left( {TMDK} \right)
\end{array} \right.\\
\Rightarrow x = 2
\end{array}\]