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Điều kiện $x \ge 1$ hoặc $x=-1.$
PT
$\Leftrightarrow \sqrt{2(x+1)(x+3)}+\sqrt{x^2-1}=2x+2$
$\Leftrightarrow ( \sqrt{2(x+1)(x+3)}+\sqrt{x^2-1})^2=(2x+2)^2$
$\Leftrightarrow 3x^2+8x+5+2\sqrt{2(x+1)(x+3)}.\sqrt{x^2-1}=4x^2+8x+4$
$\Leftrightarrow 2\sqrt{2(x+1)(x+3)}.\sqrt{x^2-1}=x^2-1$
$\Leftrightarrow 8(x+1)(x+3)(x^2-1)=(x^2-1)^2$
$\Leftrightarrow \left[ {\begin{matrix} x^2-1=0\\ x+1=0\\8(x+3)=x-1 \end{matrix}} \right.$
$\Leftrightarrow \left[ {\begin{matrix} x=1\\ x=-1\\x=-25/7 \text{(loại)} \end{matrix}} \right.$
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