Sửa đề: \(x=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)
=>\(x^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{4^2-\left(10+2\sqrt5\right)}\)
=>\(x^2=8+2\cdot\sqrt{6-2\sqrt5}=8+2\cdot\sqrt{\left(\sqrt5-1\right)^2}\)
=>\(x^2=8+2\left(\sqrt5-1\right)=6+2\sqrt5\)
=>\(x^2=\left(\sqrt5+1\right)^2\)
=>\(x=\sqrt5+1\)
\(P=\frac{x^4-4x^3+x^2+6x+12}{x^2-2x+12}\)
\(=\frac{x^4-2x^3+12x^2-2x^3+4x^2-24x-15x^2+30x-180+192}{x^2-2x+12}\)
\(=x^2-2x-15+\frac{192}{x^2-2x+12}\)
\(x=\sqrt5+1\)
=>\(x^2-2x=x\left(x-2\right)=\left(\sqrt5+1\right)\left(\sqrt5+1-2\right)=\left(\sqrt5+1\right)\left(\sqrt5-1\right)=5-1=4\)
\(P=x^2-2x-15+\frac{192}{x^2-2x+12}\)
\(=4-15+\frac{192}{4+12}\)
\(=-11+\frac{192}{16}=-11+12=1\)
