Question

[Jonh Capricorn]

\(P\left(x\right)=\sqrt[3]{\sqrt{x+8}\left(x^4+8x^3+12x\right)+6x^3+48x^2+8}\)

đặt \(A=\sqrt{x+8}\left(x^4+8x^3+12x\right)+6x^3+48x^2+8\)

          \(=\sqrt{x+8}\left(x^4+8x^3\right)+6x^2\left(x+8\right)+12x\sqrt{x+8}+8\)

            \(=\sqrt{\left(x+8\right)^3}x^3+3\sqrt{\left(x+8\right)^2}x^22+3\sqrt{\left(x+8\right)}x4+8\)

            \(=\left(x\sqrt{x+8}+2\right)^3\)

\(\Rightarrow P\left(x\right)=x\sqrt{x+8}+2\)

Answer 1

[Không Tên]

\(P\left(x\right)=\sqrt[3]{\sqrt{x+8}.\left[x^3\left(x+8\right)+12x\right]+6x^2\left(x+8\right)+8}\)

Đặt:  \(\sqrt{x+8}=a>0\) =>  \(x+8=a^2\)

Khi đó ta có:

\(P\left(x\right)=\sqrt[3]{a\left(x^3a^2+12x\right)+6x^2a^2+8}\)

\(=\sqrt[3]{x^3a^3+12xa+6x^2a^2+2}\)

\(=\sqrt[3]{\left(ax+2\right)^3}\)

\(=ax+2\)

\(=x\sqrt{x+8}+2\)