Giải pt: `\sqrt{x}+\sqrt{x+7}+2\sqrt{x^2+7x}+2x=35` 31/08/2021 Bởi Gianna Giải pt: `\sqrt{x}+\sqrt{x+7}+2\sqrt{x^2+7x}+2x=35`
Đáp án: \(x = \dfrac{{841}}{{144}}\) Giải thích các bước giải: \(\begin{array}{l}DK:x \ge 0\\Đặt:\sqrt x + \sqrt {x + 7} = t\\ \to x + 2\sqrt {{x^2} + 7x} + x + 7 = {t^2}\\ \to 2x + 2\sqrt {{x^2} + 7x} + 7 = {t^2}\\ \to 2\sqrt {{x^2} + 7x} = {t^2} – 2x – 7\\Pt \to t + {t^2} – 2x – 7 + 2x = 35\\ \to {t^2} + t – 42 = 0\\ \to \left( {t – 6} \right)\left( {t + 7} \right) = 0\\ \to \left[ \begin{array}{l}t = 6\\t = – 7\left( l \right)\end{array} \right.\\ \to \sqrt x + \sqrt {x + 7} = 6\\ \to \sqrt {x + 7} = 6 – \sqrt x \\ \to x + 7 = 36 – 12\sqrt x + x\left( {DK:36 \ge x} \right)\\ \to 12\sqrt x = 29\\ \to \sqrt x = \dfrac{{29}}{{12}}\\ \to x = \dfrac{{841}}{{144}}\end{array}\) Bình luận
Đáp án:
\(x = \dfrac{{841}}{{144}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0\\
Đặt:\sqrt x + \sqrt {x + 7} = t\\
\to x + 2\sqrt {{x^2} + 7x} + x + 7 = {t^2}\\
\to 2x + 2\sqrt {{x^2} + 7x} + 7 = {t^2}\\
\to 2\sqrt {{x^2} + 7x} = {t^2} – 2x – 7\\
Pt \to t + {t^2} – 2x – 7 + 2x = 35\\
\to {t^2} + t – 42 = 0\\
\to \left( {t – 6} \right)\left( {t + 7} \right) = 0\\
\to \left[ \begin{array}{l}
t = 6\\
t = – 7\left( l \right)
\end{array} \right.\\
\to \sqrt x + \sqrt {x + 7} = 6\\
\to \sqrt {x + 7} = 6 – \sqrt x \\
\to x + 7 = 36 – 12\sqrt x + x\left( {DK:36 \ge x} \right)\\
\to 12\sqrt x = 29\\
\to \sqrt x = \dfrac{{29}}{{12}}\\
\to x = \dfrac{{841}}{{144}}
\end{array}\)