Question

Giải PT: \(\left(x+1\right).\sqrt{x^2-2x+3}=x^2+1\)

Answer 1

\(\left(x+1\right)\sqrt{x^2-2x+3}=x^2+1\)

\(\Leftrightarrow\left(x+1\right)\sqrt{x^2-2x+3}-2\left(x+1\right)-\left(x^2-2x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(\sqrt{x^2-2x+3}-2\right)-\left(x^2-2x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\dfrac{x^2-2x+3-2^2}{\sqrt{x^2-2x+3}+2}-\left(x^2-2x-1\right)=0\)

\(\Leftrightarrow\left(x^2-2x-1\right)\left(\dfrac{x+1}{\sqrt{x^2-2x+3}+2}-1\right)=0\)

\(\Leftrightarrow x^2-2x-1=0\)

(vì \(\sqrt{x^2-2x+3}>\sqrt{x^2-2x+1}=\left|x-1\right|\ge x-1=x+1-2\)

\(\Leftrightarrow\sqrt{x^2-2x+3}+2>x+1\Leftrightarrow\dfrac{x+1}{\sqrt{x^2-2x+3}+2}< 1\))

\(\Leftrightarrow x=1\pm\sqrt{2}\).