Question

Giải phương trình:

\(\sqrt{x-2}-\sqrt{x-3}=\sqrt{x-4}\)

Answer 1

ĐKXĐ: \(x\ge\)4

Đặt x - 3 = t ( t \(\ge\) 0)

\(\Rightarrow Pt=\sqrt{t+1}-\sqrt{t}=\sqrt{t-1}\)

\(\Leftrightarrow\sqrt{t+1}=\sqrt{t-1}+\sqrt{t}\)

\(\Leftrightarrow t+1=t-1+t+2\sqrt{t^2-t}\)

\(\Leftrightarrow\)\(2-t=2\sqrt{t^2-t}\)

\(\Leftrightarrow4-4t+t^2=4t^2-4t\) ( t \(\le2\))

\(\Leftrightarrow3t^2=4\)

\(\Leftrightarrow t^2=\dfrac{4}{3}\)

\(\Leftrightarrow t=\pm\dfrac{2\sqrt{3}}{3}\)

\(\Leftrightarrow t=\dfrac{2\sqrt{3}}{3}\) (Vì 0 \(\le\) t \(\le\) 2)

\(\Rightarrow x-3=\dfrac{2\sqrt{3}}{3}\)

\(\Leftrightarrow x=\dfrac{9+2\sqrt{3}}{3}\) (thỏa mãn)

Vậy \(x=\dfrac{9+2\sqrt{3}}{3}\)