Đáp án:
\(x = \dfrac{{841}}{{144}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0\\
Đặt:\sqrt x + \sqrt {x + 7} = t\\
\to x + 2\sqrt {{x^2} + 7x} + x + 7 = {t^2}\\
\to 2x + 2\sqrt {{x^2} + 7x} + 7 = {t^2}\\
\to 2\sqrt {{x^2} + 7x} = {t^2} - 2x - 7\\
Pt \to t + {t^2} - 2x - 7 + 2x = 35\\
\to {t^2} + t - 42 = 0\\
\to \left( {t - 6} \right)\left( {t + 7} \right) = 0\\
\to \left[ \begin{array}{l}
t = 6\\
t = - 7\left( l \right)
\end{array} \right.\\
\to \sqrt x + \sqrt {x + 7} = 6\\
\to \sqrt {x + 7} = 6 - \sqrt x \\
\to x + 7 = 36 - 12\sqrt x + x\left( {DK:36 \ge x} \right)\\
\to 12\sqrt x = 29\\
\to \sqrt x = \dfrac{{29}}{{12}}\\
\to x = \dfrac{{841}}{{144}}
\end{array}\)
