Giúp mình vs ruiiriirriri

Đáp án:

\[\left[ \begin{array}{l}
x = 1\\
x = \dfrac{5}{3}
\end{array} \right.\]

Giải thích các bước giải:

ĐKXĐ:  \(\left\{ \begin{array}{l}
{x^2} – 1 \ge 0\\
{x^2} + 2\sqrt {{x^2} – 1}  \ge 0
\end{array} \right. \Leftrightarrow {x^2} – 1 \ge 0 \Leftrightarrow {x^2} \ge 1 \Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le  – 1
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
\sqrt {{x^2} + 2\sqrt {{x^2} – 1} }  = 2x – 1\\
 \Leftrightarrow \sqrt {\left( {{x^2} – 1} \right) + 2\sqrt {{x^2} – 1}  + 1}  = 2x – 1\\
 \Leftrightarrow \sqrt {{{\sqrt {{x^2} – 1} }^2} + 2.\sqrt {{x^2} – 1} .1 + {1^2}}  = 2x – 1\\
 \Leftrightarrow \sqrt {{{\left( {\sqrt {{x^2} – 1}  + 1} \right)}^2}}  = 2x – 1\\
 \Leftrightarrow \sqrt {{x^2} – 1}  + 1 = 2x – 1\\
 \Leftrightarrow \sqrt {{x^2} – 1}  = 2x – 2\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x – 2 \ge 0\\
{x^2} – 1 = {\left( {2x – 2} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left( {x – 1} \right)\left( {x + 1} \right) = 4.{\left( {x – 1} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x – 1 = 0\\
x + 1 = 4.\left( {x – 1} \right)
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = 1\\
x = \dfrac{5}{3}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{5}{3}
\end{array} \right.
\end{array}\)