a) căn bậc ba (6x + 2) = 8x^3 - 4x - 2 b) căn bậc bốn (17 - x) + căn bậc bốn x = 3 c) x = căn (3 - x). căn (4 - x) + căn (4 - x). căn (5 - x) + căn (5 - x). căn (3 - x) Giúp mình với, bất cứ câu nào cũng đc 5s nếu đúng

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\sqrt[3]{{6x + 2}} = 8{x^3} - 4x - 2\\
 \Leftrightarrow \sqrt[3]{{6x + 2}} - 2 = 8{x^3} - 4x - 4\\
 \Leftrightarrow \frac{{6x + 2 - 8}}{{\sqrt[3]{{{{\left( {6x + 2} \right)}^2}}} + 2\sqrt[3]{{6x + 2}} + 4}} = \left( {8{x^3} - 8{x^2}} \right) + \left( {8{x^2} - 8x} \right) + \left( {4x - 4} \right)\\
 \Leftrightarrow \frac{{6\left( {x - 1} \right)}}{{\sqrt[3]{{{{\left( {6x + 2} \right)}^2}}} + 2\sqrt[3]{{6x + 2}} + 4}} = \left( {x - 1} \right)\left( {8{x^2} + 8x + 4} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\frac{6}{{\sqrt[3]{{{{\left( {6x + 2} \right)}^2}}} + 2\sqrt[3]{{6x + 2}} + 4}} = 8{x^2} + 8x + 4\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)
\end{array} \right.\\
\sqrt[3]{{{{\left( {6x + 2} \right)}^2}}} + 2\sqrt[3]{{6x + 2}} + 4 = {\left( {\sqrt[3]{{6x + 2}} + 1} \right)^2} + 3 \ge 3 \Rightarrow \frac{6}{{\sqrt[3]{{{{\left( {6x + 2} \right)}^2}}} + 2\sqrt[3]{{6x + 2}} + 4}} \le \frac{6}{3} = 2\\
8{x^2} + 8x + 4 = 8{\left( {x + \frac{1}{2}} \right)^2} + 2 \ge 2\\
 \Rightarrow \frac{6}{{\sqrt[3]{{{{\left( {6x + 2} \right)}^2}}} + 2\sqrt[3]{{6x + 2}} + 4}} \ge 2 \ge 8{x^2} + 8x + 4\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
{\left( {\sqrt[3]{{6x + 2}} + 1} \right)^2} = 0\\
{\left( {x + \frac{1}{2}} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
6x + 2 =  - 1\\
x =  - \frac{1}{2}
\end{array} \right. \Leftrightarrow x =  - \frac{1}{2}
\end{array}\)

Vậy \(\left[ \begin{array}{l}
x = 1\\
x =  - \frac{1}{2}
\end{array} \right.\)