ĐK: \(x\ge2\)
PT trở thành:
\(\sqrt{9\left(x-2\right)}-\dfrac{1}{2}\sqrt{4\left(x-2\right)}+\sqrt{x-2}=1\\ \Leftrightarrow3\sqrt{x-2}-\sqrt{x-2}+\sqrt{x-2}=1\\ \Leftrightarrow\sqrt{x-2}\left(3-1+1\right)=1\\ \Leftrightarrow\sqrt{x-2}=\dfrac{1}{3}\\ \Leftrightarrow x-2=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\\ \Leftrightarrow x=\dfrac{19}{9}\left(tm\right)\)
√(9x - 18) - 1/2 √(4x - 8) + √(x - 2) = 1
⇔ 3√(x - 2) - √(x - 2) + √(x - 2) = 1
⇔ 3√(x - 2) = 1
⇔ √(x - 2) = 1/3 (1)
ĐKXĐ: x ≥ 2
(1) ⇔ x - 2 = 1/9
⇔ x = 1/9 + 2
⇔ x = 19/9 (nhận)
Vậy x = 19/9