a) Ta có: \(P=\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\)
\(\Rightarrow P^3=\left(\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\right)^3\)
\(=22\sqrt{2}+25-22\sqrt{2}+25-3\sqrt[3]{\left(22\sqrt{2}+25\right)\left(22\sqrt{2}-25\right)}\left(\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\right)\)
\(=50-3\sqrt[3]{968-625}.P\)
\(=50-3\sqrt[3]{343}.P\)
\(=50-21P\)
\(\Leftrightarrow P^3+21P-50=0\)
\(\Leftrightarrow\left(P-2\right)\left(P^2-2P+25\right)=0\)
\(\Leftrightarrow P-2=0\) (vì \(P^2-2P+25=\left(P^2-2P+1\right)+24=\left(P-1\right)^2+24\ge24\))
\(\Leftrightarrow P=2\)
Vậy \(P=2.\)
b) Theo giả thiết, ta có:
\(x=\sqrt[3]{3}+2\)
\(\Leftrightarrow x-2=\sqrt[3]{3}\)
\(\Rightarrow\left(x-2\right)^3=\left(\sqrt[3]{3}\right)^3\)
\(\Leftrightarrow x^3-6x^2+12x-8=3\)
\(\Leftrightarrow x^3-6x^2+12x-11=0\)
Do đó:
\(P=x^5-6x^4+12x^3-11x^2+20\)
\(=x^2\left(x^3-6x^2+12x-11\right)+20\)
\(=20\)
