$\sqrt[]{x-1}$ +$\sqrt[]{x+7}$+ $x^{2}$ -3x-2=0

Giải thích các bước giải:

\(\begin{array}{l}
\sqrt {x – 1}  + \sqrt {x + 7}  + {x^2} – 3x – 2 = 0(*)\\
Dk:x \ge 1\\
(*) \Leftrightarrow (\sqrt {x – 1}  – 1) + (\sqrt {x + 7}  – 3) + {x^2} – 3x + 2 = 0\\
 \Leftrightarrow \frac{{x – 2}}{{\sqrt {x – 1}  + 1}} + \frac{{x – 2}}{{\sqrt {x + 7}  + 3}} + (x – 2)(x – 1) = 0\\
 \Leftrightarrow (x – 2)\left( {\frac{1}{{\sqrt {x – 1}  + 1}} + \frac{1}{{\sqrt {x + 7}  + 3}} + x – 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2 \to Tm\\
\frac{1}{{\sqrt {x – 1}  + 1}} + \frac{1}{{\sqrt {x + 7}  + 3}} + x – 1 = 0(**)
\end{array} \right.\\
Ta\;co:\frac{1}{{\sqrt {x – 1}  + 1}} > 0;\frac{1}{{\sqrt {x + 7}  + 3}} > 0;x – 1 \ge \forall x \ge 1\\
 \Rightarrow (**)\;vo\;nghiem\\
Vay\;nghiem\;pt\;la\;x = 2
\end{array}\)