\(x^2+4x=\left(x+2\right)\sqrt{x^2-2x+4}\)
\(\Leftrightarrow x^2+4x=\left(\sqrt{\left(x+2\right)^2}\right)\cdot\sqrt{x^2-2x+4}\)
\(\Leftrightarrow x^2+4x=\sqrt{\left(x+2\right)^2\cdot\left(x^2-2x+4\right)}\)
\(\Leftrightarrow x^2+4x=\sqrt{\left(x^2+4x+4\right)\cdot\left(x^2-2x+4\right)}\)
\(\Leftrightarrow x^2+4x=\sqrt{x^4-2x^3+4x^2+4x^3-8x^2+16x+4x^2-8x+16}\)
\(\Leftrightarrow x^2+4x=\sqrt{x^4+2x^3-8x+16}\)
\(\Leftrightarrow\sqrt{\left(x^2+4x\right)^2}=\sqrt{x^4+2x^3-8x+16}\)
\(\rightarrow\left(x^2+4x\right)^2=x^4+2x^3-8x+16\)
\(\Leftrightarrow x^4+8x^3+16x^2=x^4+2x^3-8x+16\)
\(\Leftrightarrow6x^3+16x^2+8x-16=0\)
\(\Leftrightarrow2\cdot\left(3x^3+8x^2+4x-4\right)=0\)
\(\Rightarrow3x^3+8x^2+4x-4=0\)
Có \(\Delta=b^2-3ac=28>1\Rightarrow\) pt có 1 nghiệm duy nhất.
Đặt \(k=\dfrac{9abc-2b^3-27a^2d}{2\cdot\sqrt{\left|\Delta\right|^3}}\)
\(\Rightarrow x=\dfrac{\sqrt{\Delta}\cdot\left|k\right|}{3ak}\cdot\left(\sqrt[3]{\left|k\right|+\sqrt{k^2-1}}+\sqrt[3]{\left|k\right|-\sqrt{k^2-1}}\right)-\dfrac{b}{3a}\)
Đến đây là ok rồi nhé :) thế vào tìm x là xong.