Giúp e giải hộ bài này với ạ 2×căn(2x-5)+2×căn(3x-5)=x^2-8x+21

Đáp án:

\[\left[ \begin{array}{l}
x = 3\\
x = 7
\end{array} \right.\]

Giải thích các bước giải:

 ĐKXĐ:\(x \ge \frac{5}{2}\)

Ta có:

\(\begin{array}{l}
2\sqrt {2x – 5}  + 2\sqrt {3x – 5}  = {x^2} – 8x + 21\\
 \Leftrightarrow \left( {2\sqrt {2x – 5}  – \left( {x – 1} \right)} \right) + \left( {2\sqrt {3x – 5}  – \left( {x + 1} \right)} \right) = {x^2} – 8x + 21 – \left( {x – 1} \right) – \left( {x + 1} \right)\\
 \Leftrightarrow \frac{{4\left( {2x – 5} \right) – {{\left( {x – 1} \right)}^2}}}{{2\sqrt {2x – 5}  + x – 1}} + \frac{{4\left( {3x – 5} \right) – {{\left( {x + 1} \right)}^2}}}{{2\sqrt {3x – 5}  + x + 1}} = {x^2} – 10x + 21\\
 \Leftrightarrow \frac{{ – {x^2} + 10x – 21}}{{2\sqrt {2x – 5}  + x – 1}} + \frac{{ – {x^2} + 10x – 21}}{{2\sqrt {3x – 5}  + x + 1}} = {x^2} – 10x + 21\\
 \Leftrightarrow \left( {{x^2} – 10x + 21} \right)\left( {1 + \frac{1}{{2\sqrt {2x – 5}  + x – 1}} + \frac{1}{{2\sqrt {3x – 5}  + x + 1}}} \right) = 0\\
x \ge \frac{5}{2} \Rightarrow 1 + \frac{1}{{2\sqrt {2x – 5}  + x – 1}} + \frac{1}{{2\sqrt {3x – 5}  + x + 1}} > 0\\
 \Rightarrow {x^2} – 10x + 21 = 0\\
 \Leftrightarrow \left( {x – 3} \right)\left( {x – 7} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 7
\end{array} \right.\left( {t/m} \right)
\end{array}\)