$GPT$ $\sqrt[]{2x^2-3x+1}$ + $\sqrt[]{x^2+x-2}$ = $\sqrt[]{3x^2-4x+1}$

Đáp án:

\[x = 1\,\,\,\,hoặc\,\,\,\,\,x = \frac{{ – 3 – \sqrt {33} }}{4}\]

Giải thích các bước giải:

 ĐKXĐ:  \(\left\{ \begin{array}{l}
2{x^2} – 3x + 1 \ge 0\\
{x^2} + x – 2 \ge 0\\
3{x^2} – 4x + 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left( {x – 1} \right)\left( {2x – 1} \right) \ge 0\\
\left( {x – 1} \right)\left( {x + 2} \right) \ge 0\\
\left( {x – 1} \right)\left( {3x – 1} \right) \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 1\\
x \le \frac{1}{2}
\end{array} \right.\\
\left[ \begin{array}{l}
x \ge 1\\
x \le  – 2
\end{array} \right.\\
\left[ \begin{array}{l}
x \ge 1\\
x \le \frac{1}{3}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le  – 2
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
TH1:\,\,\,\,x \ge 1\\
\sqrt {2{x^2} – 3x + 1}  + \sqrt {{x^2} + x – 2}  = \sqrt {3{x^2} – 4x + 1} \\
 \Leftrightarrow \sqrt {\left( {2x – 1} \right)\left( {x – 1} \right)}  + \sqrt {\left( {x + 2} \right)\left( {x – 1} \right)}  = \sqrt {\left( {3x – 1} \right)\left( {x – 1} \right)} \\
 \Leftrightarrow \sqrt {x – 1} .\sqrt {2x – 1}  + \sqrt {x – 1} .\sqrt {x + 2}  – \sqrt {3x – 1} .\sqrt {x – 1}  = 0\\
 \Leftrightarrow \sqrt {x – 1} .\left( {\sqrt {2x – 1}  + \sqrt {x + 2}  – \sqrt {3x – 1} } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x – 1}  = 0\\
\sqrt {2x – 1}  + \sqrt {x + 2}  – \sqrt {3x – 1}  = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\sqrt {2x – 1}  + \sqrt {x + 2}  = \sqrt {3x – 1} 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
2x – 1 + 2.\sqrt {\left( {2x – 1} \right)\left( {x + 2} \right)}  + x + 2 = 3x – 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
3x + 2 + 2\sqrt {\left( {2x – 1} \right)\left( {x + 2} \right)}  = 3x – 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
2\sqrt {\left( {2x – 1} \right)\left( {x + 2} \right)}  =  – 3\,\,\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)
\end{array} \right.\\
 \Rightarrow x = 1\\
TH2:\,\,\,\,\,x \le  – 2\\
\sqrt {2{x^2} – 3x + 1}  + \sqrt {{x^2} + x – 2}  = \sqrt {3{x^2} – 4x + 1} \\
 \Leftrightarrow \sqrt {\left( {2x – 1} \right)\left( {x – 1} \right)}  + \sqrt {\left( {x + 2} \right)\left( {x – 1} \right)}  = \sqrt {\left( {3x – 1} \right)\left( {x – 1} \right)} \\
 \Leftrightarrow \sqrt {1 – x} .\sqrt {1 – 2x}  + \sqrt {1 – x} .\sqrt { – \left( {x + 2} \right)}  = \sqrt {1 – x} .\sqrt {1 – 3x} \\
 \Leftrightarrow \sqrt {1 – x} .\left( {\sqrt {1 – 2x}  + \sqrt { – x – 2}  – \sqrt {1 – 3x} } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {1 – x}  = 0\\
\sqrt {1 – 2x}  + \sqrt { – x – 2}  – \sqrt {1 – 3x}  = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {L,\,\,x \le  – 2} \right)\\
\sqrt {1 – 2x}  + \sqrt { – x – 2}  = \sqrt {1 – 3x} 
\end{array} \right.\\
 \Leftrightarrow \left( {1 – 2x} \right) + 2.\sqrt {\left( {1 – 2x} \right)\left( { – x – 2} \right)}  + \left( { – x – 2} \right) = 1 – 3x\\
 \Leftrightarrow 2\sqrt {\left( {1 – 2x} \right)\left( { – x – 2} \right)}  = 2\\
 \Leftrightarrow \sqrt {\left( {1 – 2x} \right)\left( { – x – 2} \right)}  = 1\\
 \Leftrightarrow 2{x^2} + 3x – 2 = 1\\
 \Leftrightarrow 2{x^2} + 3x – 3 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ – 3 – \sqrt {33} }}{4}\\
x = \frac{{ – 3 + \sqrt {33} }}{4}
\end{array} \right.\\
x \le  – 2 \Rightarrow x = \frac{{ – 3 – \sqrt {33} }}{4}
\end{array}\)

Vậy \(x = 1\,\,\,\,hoặc\,\,\,\,\,x = \frac{{ – 3 – \sqrt {33} }}{4}\)