Đáp án:
\[x = 1\]
Giải thích các bước giải:
ĐKXĐ: \(x \le 1\)
Ta có:
\(\begin{array}{l}
\sqrt[4]{{1 - x}} + \sqrt[4]{{2 - x}} = \sqrt[4]{{3 - 2x}}\\
\Leftrightarrow {\left( {\sqrt[4]{{1 - x}} + \sqrt[4]{{2 - x}}} \right)^4} = 3 - 2x\\
\Leftrightarrow 1 - x + 4.{\sqrt[4]{{1 - x}}^3}.\sqrt[4]{{2 - x}} + 6.{\sqrt[4]{{1 - x}}^2}.{\sqrt[4]{{2 - x}}^2} + 4.\sqrt[4]{{1 - x}}.{\sqrt[4]{{2 - x}}^3} + 2 - x = 3 - 2x\\
\Leftrightarrow 4.{\sqrt[4]{{1 - x}}^3}.\sqrt[4]{{2 - x}} + 6.{\sqrt[4]{{1 - x}}^2}.{\sqrt[4]{{2 - x}}^2} + 4.\sqrt[4]{{1 - x}}.{\sqrt[4]{{2 - x}}^3} = 0\\
\Leftrightarrow \sqrt[4]{{1 - x}}.\sqrt[4]{{2 - x}}.\left( {2.{{\sqrt[4]{{1 - x}}}^2} + 3.\sqrt[4]{{1 - x}}.\sqrt[4]{{2 - x}} + 2{{\sqrt[4]{{2 - x}}}^2}} \right) = 0\\
2.{\sqrt[4]{{1 - x}}^2} + 3.\sqrt[4]{{1 - x}}.\sqrt[4]{{2 - x}} + 2{\sqrt[4]{{2 - x}}^2} > 0,\,\,\,\,\,\forall x \le 1\\
\Rightarrow \sqrt[4]{{1 - x}}.\sqrt[4]{{2 - x}} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt[4]{{1 - x}} = 0\\
\sqrt[4]{{2 - x}} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {t/m} \right)\\
x = 2\left( L \right)
\end{array} \right.
\end{array}\)
Vậy \(x = 1\)
