Question

[Trúc Giang]

Tính GTBT: \(M=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)\) biết

\(x=\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\)

\(y=\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\)

Answer 1

[Lê Thị Thục Hiền]

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 \(M=-20\sqrt{2}\)

Answer 2

[missing you =]

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