\(x=1+\sqrt[3]{2}+\sqrt[3]{4}\)
\(\Leftrightarrow x-1=\sqrt[3]{2}+\sqrt[3]{4}\)
\(\Leftrightarrow\left(x-1\right)^3=\left(\sqrt[3]{2}+\sqrt[3]{4}\right)^3\)
\(\Leftrightarrow x^3-3x^2+3x-1=2+4+3\sqrt[3]{2.4}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1=6+6\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1=6+6\left(x-1\right)\)( vì \(x-1=\sqrt[3]{2}+\sqrt[3]{4}\))
\(\Leftrightarrow x^3-3x^2-3x-1=0\)
Ta có \(M=\frac{\sqrt{x^3+x^2+5x+3}-6}{\sqrt{x^3-2x^2-7x+3}}\)
\(=\frac{\sqrt{x^3-3x^2-3x-1+4x^2+8x+4}-6}{\sqrt{x^3-3x^2-3x-1+x^2-4x+4}}\)
\(=\frac{\sqrt{4x^2+8x+4}-6}{\sqrt{x^2-4x+4}}\)( vì \(x^3-3x^2-3x-1=0\))
\(=\frac{\sqrt{\left(2x+2\right)^2}-6}{\sqrt{\left(x-2\right)^2}}\)
\(=\frac{\left|2x+2\right|-6}{\left|x-2\right|}\)
Từ điều kiện đề bài \(\Rightarrow x>2\)
\(\Rightarrow M=\frac{2x+2-6}{x-2}=2\)
Vậy \(M=2\)\(\Leftrightarrow x=1+\sqrt[3]{2}+\sqrt[3]{4}\)
