thay \(x=3-2\sqrt{2}\) vào P ta có:
\(\dfrac{x+8}{\sqrt{x}+1}=\dfrac{3-2\sqrt{2}+8}{\sqrt{3-2\sqrt{2}}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}-1+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}\)
\(b,x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
Thay vào P, ta được:
\(P=\dfrac{3-2\sqrt{2}+8}{\sqrt{\left(\sqrt{2}-1\right)^2}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}=\dfrac{11\sqrt{2}-4}{2}\)
b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{3+2\sqrt{2}+8}{2+\sqrt{2}}=\dfrac{18-7\sqrt{2}}{2}\)