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Trả lời:
$a, \sqrt{x^2-x+2}+x=6$
$⇔\sqrt{x^2-x+2}=6-x$
$⇔x^2-x+2=36-12x+x^2 (ĐK: x \leq 6)$
$⇔11x=34$
$⇔x=\dfrac{34}{11}$
Vậy $S=\{\dfrac{34}{11}\}$.
$b, \sqrt{2x^2-2x+1}-2=x$
$⇔\sqrt{2x^2-2x+1}=x+2$
$⇔2x^2-2x+1=x^2+4x+4$ $(ĐK: x \geq -2)$
$⇔x^2-6x-3=0$
$⇔x=3±2\sqrt{3}$
Vậy $S=\{3±2\sqrt{3}\}$.