\(\left(x+1\right)\sqrt{x^2-2x+3}=x^2+1\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2-2x+3\right)=\left(x^2+1\right)^2\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(x^2-2x+3\right)=x^4+2x^2+1\)
\(\Leftrightarrow x^4-2x^3+3x^2+2x^3-4x^2+6x+x^2-2x+3=x^4+2x^2+1\)
\(\Leftrightarrow x^4+4x+3=x^4+2x^2+1\)
\(\Leftrightarrow x^4+4x+3-x^4-2x^2-1=0\)
\(\Leftrightarrow-2x^2+4x+2=0\)
\(\Leftrightarrow x^2-2x-1=0\)
\(\Delta=\left(-2\right)^2+4.1=4+4=8>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{2+\sqrt{8}}{2}=1+\sqrt{2}\\x_2=\dfrac{2-\sqrt{8}}{2}=1-\sqrt{2}\end{matrix}\right.\)
Vậy................
\(\left(x+1\right)\sqrt{x^2-2x+3}=x^2+1\) Điều kiện : x > -1
\(\Leftrightarrow\left(x+1\right)^2\left(x^2-2x+3\right)=\left(x^2+1\right)^2\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(x^2-2x+3\right)=x^4+2x^2+1\)(1)
\(\Leftrightarrow x^4-2x^3+3x^2+2x^3-4x^2+6x+x^2-2x+3=x^4+2x^2+1\)
\(\Leftrightarrow x^2-2x-1=0\Rightarrow\left\{{}\begin{matrix}x=1-\sqrt{2}\\x=1+\sqrt{2}\end{matrix}\right.\)(thỏa mãn điều kiện )
vậy x=\(\left\{{}\begin{matrix}1-\sqrt{2}\\1+\sqrt{2}\end{matrix}\right.\) là nghiệm của phương trình (1)
