Question

giải pt x\(^2\)- 1 = 2\(\sqrt{2x+1}\)

Answer 1

\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow x^4-2x^2+1=8x+4\\ \Leftrightarrow x^4-2x^2-8x-3=0\\ \Leftrightarrow x^4-2x^3-x^2+2x^3-4x^2-2x+3x^2-6x-3=0^4xx\\ \Leftrightarrow\\ \Leftrightarrow\left(x^2-2x-1\right)\left(x^2+2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-2x-1=0\\x^2+2x+3=0\left(\text{vô nghiệm}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{2}\left(tm\right)\\x=1-\sqrt{2}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1+\sqrt{2}\)