Đặt \(\sqrt[3]{1+\frac{\sqrt{84}}{9}}=a;\sqrt[3]{1-\frac{\sqrt{84}}{9}}=b\Rightarrow x=a+b;a^3+b^3=2;ab=-\frac{1}{3}\)

Ta có:\(x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

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

\(\Leftrightarrow x=1\).Vì \(x^2+x+2=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\)

Từ đó suy ra điều phải chứng minh

~~~~~~~~~~~ Chúc bạn hok tốt~~~~~~~~~~~~

Tính giá trị của biểu thức \(P=x^3+y^3-3\left(x+y\right)+2004\)

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

Đặt \(\sqrt[3]{1+\frac{\sqrt{84}}{9}}=a\);\(\sqrt[3]{1-\frac{\sqrt{84}}{9}}=b\)

\(\Rightarrow x=a+b;a^3+b^3=2;ab=-\frac{1}{3}\)

Ta có: \(x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

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

\(\Leftrightarrow x=1\).vì \(x^2+x+2=0=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\)

=> đpcm

P/s tham khảo

Ta có : \(x=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)

\(\Leftrightarrow x^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)

\(\Leftrightarrow x^3=1+\frac{\sqrt{84}}{9}+1-\frac{\sqrt{84}}{9}+3.\sqrt[3]{1+\frac{\sqrt{84}}{9}}.\sqrt[3]{1-\frac{\sqrt{84}}{9}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}^3\right)\)

\(\Leftrightarrow x^3=2+3.\sqrt[3]{1^2-\frac{84}{81}}.x\Leftrightarrow x^3=2-x\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x^2+x+2=0\end{array}\right.\)

Vì \(x^2+x+2=\left(x^2+x+\frac{1}{4}\right)+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\) nên pt này vô nghiệm.
Vậy x - 1 = 0 => x = 1

Vậy x có giá trị là số nguyên.

Đặt \(P=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)

\(P^3=1+\frac{\sqrt{84}}{9}+1-\frac{\sqrt{84}}{9}+3\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)\left(1-\frac{\sqrt{84}}{9}\right)}\cdot P\)

\(P^3=2+3\sqrt[3]{1-\frac{84}{81}}\cdot P\)

\(P^3=2+3\sqrt[3]{\frac{-1}{27}}\cdot P\)

\(P^3=2+3\cdot\frac{-1}{3}\cdot P\)

\(P^3=2-P\)

\(\Leftrightarrow P^3+P-2=0\)

\(\Leftrightarrow P^3-P^2+P^2-P+2P-2=0\)

\(\Leftrightarrow P^2\left(P-1\right)+P\left(P-1\right)+2\left(P-1\right)=0\)

\(\Leftrightarrow\left(P-1\right)\left(P^2+P+2\right)=0\)

Do \(P^2+P+2>0\forall P\)

Do đó \(P-1=0\Leftrightarrow P=1\)

Vậy \(P=1\) là một số nguyên ( đpcm )

Đặt \(A=\sqrt[3]{1+\dfrac{\sqrt{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt{84}}{9}}\)

\(\Rightarrow A^3=1+\dfrac{\sqrt{84}}{9}+1-\dfrac{\sqrt{84}}{9}+3A.\sqrt[3]{1+\dfrac{\sqrt{84}}{9}}.\sqrt[3]{1-\dfrac{\sqrt{84}}{9}}\)

\(\Leftrightarrow A^3=2+3A.\sqrt[3]{1-\dfrac{84}{81}}\)

\(\Leftrightarrow A^3=2+3A.\sqrt[3]{-\dfrac{3}{81}}=2+3A.\sqrt[3]{-\dfrac{1}{27}}\)

\(\Leftrightarrow A^3=2-A\)

\(\Leftrightarrow A^3+A-2=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+2\right)=0\)

Dể thấy \(A^2+A+2=\left(A+\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)

\(\Rightarrow A-1=0\Leftrightarrow A=1\)

Vậy \(\sqrt[3]{1+\dfrac{\sqrt{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt{84}}{9}}\) là số nguyên (đpcm)

\(x^3=1+\frac{\sqrt{84}}{9}+1-\frac{\sqrt{84}}{9}+3x\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)\left(1-\frac{\sqrt{84}}{9}\right)}\)

     \(=2+3x\sqrt[3]{1-\frac{84}{81}}\)

     \(=2+3x\sqrt[3]{-\frac{1}{27}}\)

    \(=2-x\)

\(\Rightarrow x^3+x-2=0\)

\(\Leftrightarrow x=1\)