Đáp án:
$S=\{3\}$
Giải thích các bước giải:
$79.|x^2+1|-\sqrt[]{x^2-4x+4}=3x$
$⇔|x^2+1|-3x=\sqrt[]{x^2-2.2.x+2^2}$
$⇔x^2+1-3x=\sqrt[]{(x-2)^2}$
$⇔x^2-3x+1=|x-2|$
Điều kiện :$x-2≥0⇔x≥2⇒|x-2|=x-2$
$x-2<0⇔x<2⇒|x-2|=2-x$
$⇔$\(\left[ \begin{array}{l}x^2-3x+1=x-2\\x^2-3x+1=2-x\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l} x^2-4x+3=0\\x^2-2x-1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l} x^2-3x-x+3=0\\x^2-2x+1-2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l} (x-3)x-(x-3)=0\\(x-1)^2=2\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l} (x-3)(x-1)=0\\(x-1)=±\sqrt[]{2}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l} x=3\\x=1\end{array} \right.\)
$⇔$$\left[ \begin{array}{l}\left[ \begin{array}{l} x=3(n)\\x=1(l)\end{array} \right. \\\left[ \begin{array}{l} x=1+\sqrt[]{2}(l)\\x=1-\sqrt[]{2}(l)\end{array} \right. \end{array} \right. $
Vậy $S=\{3\}$
