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Có \(x^3=3+2\sqrt{2}-3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)-\left(3-2\sqrt{2}\right)\)
\(\Leftrightarrow x^3=4\sqrt{2}-3x\) \(\Leftrightarrow x^3+3x=4\sqrt{2}\) (1)
Có \(y^3=17+12\sqrt{2}-3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\right)-\left(17-12\sqrt{2}\right)\)
\(\Leftrightarrow y^3=24\sqrt{2}-3y\) \(\Leftrightarrow y^3+3y=24\sqrt{2}\) (2)
Từ (1) (2)\(\Rightarrow x^3+3x-y^3-3y=-20\sqrt{2}\)
Có \(M=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)=\left(x-y\right)\left[\left(x-y\right)^2+3\left(xy+1\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2+3\right)=x^3-y^3+3\left(x-y\right)=-20\sqrt{2}\)
Vậy \(M=-20\sqrt{2}\)
theo bài ra
\(x=\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\)
\(=>x^3=\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(x^3=4\sqrt{2}-3\left[\left(\sqrt[3]{3+2\sqrt{2}}\right)\left(\sqrt[3]{3-2\sqrt{2}}\right)\right]\left[\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right]\)
\(x^3=4\sqrt{2}-3\left[\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\right].x\)
\(x^3=4\sqrt{2}-3.\left[\sqrt[3]{9-\left(2\sqrt{2}\right)^2}\right]x\)
\(x^3=4\sqrt{2}-3.1x\)
\(x^3=4\sqrt{2}-3x\)
\(< =>x^3+3x-4\sqrt{2}=0\)
rồi làm y tương tự rồi thế vào M là ra
\(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
\(\Rightarrow x^3=3+2\sqrt{2}+3-2\sqrt{2}+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)
\(=6+3\sqrt[3]{9-8}.x=6+3x\)
\(\Rightarrow x^3-3x=6\)
\(y=\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\)
\(\Rightarrow y^3=17+12\sqrt{2}+17-12\sqrt{2}+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\right)\)
\(=34+3\sqrt[3]{289-288}.y=34+3y\)
\(\Rightarrow y^3-3y=34\)
\(P=x^3+y^3-3\left(x+y\right)+2009=\left(x^3-3x\right)+\left(y^3-3y\right)+2009\)
\(=6+34+2009=2049\)
làm ra chưa chỉ với bạn