Căn thức đằng sau là căn bậc 2 hay căn bậc 3 bạn?

Căn bậc 2 thì x nó vô tỉ chứ hữu tỉ làm sao được

\(x^3=2a+3x.\sqrt[3]{a^2-\frac{\left(a+1\right)^2}{9}\left(\frac{8a-1}{3}\right)}\)

\(x^3=2a+3x\sqrt[3]{\frac{1-6a+12a^2-8a^3}{27}}\)

\(x^3=2a+3x\sqrt[3]{\left(\frac{1-2a}{3}\right)^3}\)

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

\(x^3-x+2ax-2a=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+x+2a=0\left(1\right)\end{matrix}\right.\)

Xét (1): \(x^2+x+\frac{1}{4}+2a-\frac{1}{4}=0\Rightarrow\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)=0\)

Do \(a>\frac{1}{8}\Rightarrow\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)>0\)

\(\Rightarrow\left(1\right)\) vô nghiệm \(\Rightarrow x=1\) hay x nguyên dương

\(x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\)

\(\Leftrightarrow x^3=\left(\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\right)^3\)

\(\Leftrightarrow x^3=2a+3.\sqrt[3]{a^2-\frac{\left(a+1\right)^2}{9}.\frac{8a-1}{3}}.\left(\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\right)\)

\(\Leftrightarrow x^3=2a+3.\sqrt[3]{\frac{-8a^3+12a^2+6a-1}{27}}.x\)

\(\Leftrightarrow x^3=2a+3.\sqrt[3]{-\left(\frac{2a-1}{3}\right)^3}.x\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x^2+x+2a=0\end{array}\right.\) . Ta có : \(x^2+x+2a=\left(x^2+x+\frac{1}{4}\right)+2\left(a-\frac{1}{8}\right)=\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)\ge2\left(a-\frac{1}{8}\right)\)

Vì \(a>\frac{1}{8}\Rightarrow x^2+x+2a>0\) => vô nghiệm.

Vậy x = 1 => x là số tự nhiên.

a) Ta có : \(x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\)

\(\Rightarrow x^3=2a+3.\sqrt[3]{a^2-\left(\frac{a+1}{3}\right)^2\left(\frac{8a-1}{3}\right)}.x\)

\(=2a+3\sqrt[3]{a^2-\frac{\left(a^2+2a+1\right)\left(8a-1\right)}{27}}.x\)

\(=2a+3\sqrt[3]{\frac{27a^2-\left(8a^3+15a^2+6a-1\right)}{27}}.x\)

\(=2a+3\sqrt[3]{\frac{-8a^3+12a^2-6a+1}{27}}.x\)

\(=2a+3x.\sqrt[3]{\frac{\left(1-2a\right)^3}{3^3}}=2a+3x.\frac{1-2a}{3}=2a+x\left(1-2a\right)\)

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

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+2a\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+x+2a=0\end{cases}}\)

Vì \(a>\frac{1}{8}\) nên \(x^2+x+2a>0\Rightarrow\)vô nghiệm.

Vậy x - 1 = 0  => x = 1 thoả mãn x là số nguyên dương.

b) \(\sqrt[3]{x+24}+\sqrt{12-x}=6\) (ĐKXĐ : \(x\le12\))

\(\Leftrightarrow\sqrt[3]{x+24}=6-\sqrt{12-x}\Leftrightarrow x+24=\left(6-\sqrt{12-x}\right)^3\)

\(\Leftrightarrow x+24=6^3-3.6^2.\sqrt{12-x}+3.6.\left(12-x\right)-\left(\sqrt{12-x}\right)^3\)

\(\Leftrightarrow x+24=216-108\sqrt{12-x}+216-18x-\sqrt{12-x}^3\)

\(\Leftrightarrow-19\left(12-x\right)+108\sqrt{12-x}+\sqrt{12-x}^3-180=0\)

 Đặt \(y=\sqrt{12-x},y\ge0\) . Phương trình trên tương đương với : 

\(-19y^2+108y+y^3-180=0\Leftrightarrow\left(y-10\right)\left(y-6\right)\left(y-3\right)=0\)

=> y = 10 (TM) hoặc y = 6 (TM) hoặc y = 3 (TM)

Vậy tập nghiệm của phương trình : \(S=\left\{-88;-24;3\right\}\)