ĐK: \(x\ge-2\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{x^2-2x+4}=b\ge0\end{matrix}\right.\Leftrightarrow a^2+b^2=x^2-x+6\), PTTT:
\(5ab=2\left(a^2+b^2\right)\\ \Leftrightarrow2a^2-5ab+2b^2=0\\ \Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)
Với \(a=2b\Leftrightarrow x+2=4\left(x^2-2x+4\right)\)
\(\Leftrightarrow4x^2-8x+16=x+2\\ \Leftrightarrow4x^2-9x+14=0\\ \Delta=81-224< 0\\ \Leftrightarrow x\in\varnothing\)
Với \(2a=b\Leftrightarrow4\left(x+2\right)=x^2-2x+4\)
\(\Leftrightarrow4x+8=x^2-2x+4\\ \Leftrightarrow x^2-6x-4=0\\ \Delta=36+16=52\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{13}\left(tm\right)\\x=3-\sqrt{13}\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm \(x=3\pm\sqrt{13}\)
Lời giải:
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow 5\sqrt{(x+2)(x^2-2x+4)}=2(x^2-x+6)$
Đặt $\sqrt{x+2}=a; \sqrt{x^2-2x+4}=b(a,b\geq 0)$ thì pt trở thành:
$5ab=2(a^2+b^2)$
$\Leftrightarrow 2a^2-5ab+2b^2=0$
$\Leftrightarrow (2a-b)(a-2b)=0$
$\Rightarrow 2a=b$ hoặc $a=2b$
Nếu $2a=b\Leftrightarrow 4a^2=b^2$
$\Leftrightarrow 4(x+2)=x^2-2x+4$
$\Leftrightarrow x=3\pm \sqrt{13}$ (tm)
Nếu $a=2b\Leftrightarrow a^2=4b^2$
$\Leftrightarrow x+2=4(x^2-x+6)$
$\Leftrightarrow 4x^2-5x+22=0$ (dễ thấy pt này vô nghiệm)
