1) ĐK: 8+2x−x2≥0⇔−2≤x≤4PT⇔(2√x2+3−4)+(3−√8+2x−x2)−(x−1)=0⇔4(x2−1)2√x2+3+4+(x−1)23+√8+2x−x2−(x−1)=0⇔(x−1)(42√x2+3+4+x−13+√8+2x−x2−1)=0⇔x=1 (t/m).
1) ĐK: 8+2x−x2≥0⇔−2≤x≤4PT⇔(2√x2+3−4)+(3−√8+2x−x2)−(x−1)=0⇔4(x2−1)2√x2+3+4+(x−1)23+√8+2x−x2−(x−1)=0$\Leftrightarrow (x-1)(\frac{4(x+1)}{2\sqrt{x^{2}+3}+4}+\frac{x-1}{3+\sqrt{8+2x-x^{2}}}-1)=0$$\Leftrightarrow x=1 $ (t/m).