điều kiện chung: \(x\ge\dfrac{5}{2}\)
phương trình \(\Leftrightarrow2\sqrt{2x-5}+2\sqrt{3x-5}-x^2+8x-21=0\)
\(\Leftrightarrow2\sqrt{2x-5}-2+2\sqrt{3x-5}-4-x^2+8x-15=0\)
\(\Leftrightarrow2\left(\sqrt{2x-5}-1+\sqrt{3x-5}-2\right)-x^2+8x-15=0\)
\(\Leftrightarrow2\left(\dfrac{2x-5-1}{\sqrt{2x-5}+1}+\dfrac{3x-5-4}{\sqrt{3x-5}+2}\right)-\left(x-5\right)\left(x-3\right)=0\)
\(\Leftrightarrow2\left[\dfrac{2\left(x-3\right)}{\sqrt{2x-5}+1}+\dfrac{3\left(x-3\right)}{\sqrt{3x-5}+2}\right]-\left(x-5\right)\left(x-3\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(\dfrac{2}{\sqrt{2x-5}+1}+\dfrac{3}{\sqrt{3x-5}+2}-x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(n\right)\\\dfrac{2}{\sqrt{2x-5}+1}+\dfrac{3}{\sqrt{3x-5}+2}-x+5=0\left(1\right)\end{matrix}\right.\)
(1)tới đây mình bí rồi huhu :((( cho mình xin lỗi
