√x+3−4√x−1+√x+8−6√x−1=1;x≥1 √(x−1)−2×2√x−1+4+√(x−1)−6√x−1+9=1 |√x−1−2|+|√x−1−3|=1 Case - I √x−1−2+√x−1−3=1 (x ≥ 10) 2√x−1=6 x=10 Case -II √x−1−2−√x−1+3=1 (5 ≤ x ≤ 10) Case -III −√x−1−2−√x−1+3=1 (1 ≤ x ≤ 5) 2√x−1=4 x=5