Question

Cho \(x=1+\sqrt[3]{2}+\sqrt[3]{4}\)

Tính \(M=\dfrac{\sqrt{x^3+x^2+5x+3}-6}{\sqrt{x^3-2x^2-7x+3}}\)

Answer 1

\(x=1+1.\sqrt[3]{2}+\sqrt[3]{2}^2=\dfrac{\sqrt[3]{2}^3-1^3}{\sqrt[3]{2}-1}=\dfrac{1}{\sqrt[3]{2}-1}\)

\(\Leftrightarrow\dfrac{1}{x}+1=\sqrt[3]{2}\)

\(\Leftrightarrow\left(x+1\right)^3=2x^3\Leftrightarrow x^3-3x^2-3x-1=0\).

Do đó \(M=\dfrac{\sqrt{x^3+x^2+5x+3}-6}{\sqrt{x^3-2x^2-7x+3}}\)

\(M=\dfrac{\sqrt{\left(x^3-3x^2-3x-1\right)+\left(4x^2+8x+4\right)}-6}{\sqrt{\left(x^3-3x^2-3x-1\right)+\left(x^2-4x+4\right)}}\)

\(M=\dfrac{\sqrt{\left(2x+2\right)^2}-6}{\sqrt{\left(x-2\right)^2}}=\dfrac{2x+2-6}{x-2}=2\). (Do \(x>2\))