Giải các phương trình sau
a) |x2 – 2x – 3| = 2x+2
b) \(\sqrt {{x^2} - 4} = 2\left( {x - \sqrt 3 } \right)\)
a) Điều kiện: x ≥−1. Ta có:
\(\begin{array}{*{20}{l}}
{\left| {{x^2} - 2x - 3} \right| = 2x + 2}\\
{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{x^2} - 2x - 3 = 2x + 2}\\
{{x^2} - 2x - 3 = - 2x - 2}
\end{array}} \right.}\\
\begin{array}{l}
\Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{x^2} - 4x - 5 = 0}\\
{{x^2} - 1 = 0}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = - 1,x = 5}\\
{x = \pm }
\end{array}} \right.\left( n \right)
\end{array}
\end{array}\)
Vậy S = {-1, 1, 5}
b) Ta có:
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sqrt {{x^2} - 4} = 2\left( {x - \sqrt 3 } \right)\\
\Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x \ge \sqrt 3 }\\
{3{x^2} - 8\sqrt 3 x + 16 = 0}
\end{array}} \right.
\end{array}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x \ge \sqrt 3 }\\
{x = \frac{{4\sqrt 3 }}{3}}
\end{array}} \right.}
\end{array}\)
Vậy \(S = \left\{ {\frac{{4\sqrt 3 }}{3}} \right\}\)
-- Mod Toán 10
Copyright © 2021 HOCTAP247