Question

Answer 1

Lời giải:
ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow 2(x^2+2)=5\sqrt{(x+1)(x^2-x+1)}$

Đặt $\sqrt{x+1}=a; \sqrt{x^2-x+1}=b, (a,b\geq 0)$

PT trở thành:

$2(a^2+b^2)=5ab$

$\Leftrightarrow 2a^2-5ab+2b^2=0$

$\Leftrightarrow (a-2b)(2a-b)=0$

$\Rightarrow a=2b$ hoặc $2a=b$

Nếu $a=2b$

$\Leftrightarrow \sqrt{x+1}=2\sqrt{x^2-x+1}$

$\Leftrightarrow x+1=4(x^2-x+1)$

$\Leftrightarrow 4x^2-5x+3=0$

$\Leftrightarrow (2x-1,25)^2=\frac{-23}{16}< 0$ (vô lý- loại)

Nếu $2a=b$

$\Leftrightarrow 4a^2=b^2\Leftrightarrow 4(x+1)=x^2-x+1$

$\Leftrightarrow x^2-5x-3=0$

$\Leftrightarrow x=\frac{5\pm \sqrt{37}}{2}$ (tm)