`b) x - y + 1 = 2sqrt{x - y} - sqrt{x - 2}`
`ĐKXĐ: x - y >= 0, x - 2 >= 0`
`<=> x >= y, x >= 2`
Phương trình tương đương:
`<=> x - y - 2sqrt{x - y} + 1 + sqrt{x - 2} = 0`
`<=> (sqrt{x - y})^2 - 2sqrt{x - y} + 1 + sqrt{x - 2} = 0`
`<=> (sqrt{x - y} - 1)^2 + sqrt{x - 2} = 0`
Với mọi `x, y` ta có:
\(\left\{\begin{matrix}(\sqrt{x - y} - 1)^2 \ge 0\\\sqrt{x - 2} \ge 0\end{matrix}\right.\)
`<=> (sqrt{x - y} - 1)^2 + sqrt{x - 2} >= 0`
Dấu "`=`" xảy ra
`<=>` \(\left\{\begin{matrix}\sqrt{x - y} - 1 = 0\\\sqrt{x - 2} = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}\sqrt{x - y} = 1\\x - 2 = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}x - y = 1\\x = 2\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}y = 1\\x = 2\end{matrix}\right.\)
Vậy `x = 2; y = 1`
