ĐK: x ≥ \(\frac{3}{4}\)
2x + 3y + 4z + 77 = 2( \(3\sqrt{2x-1}+5\sqrt{3y-2}+7\sqrt{4x-3}\))
⇔2x +3y +4z +77= \(6\sqrt{2x-1}+10\sqrt{3y-2}+14\sqrt{4x-3}\)
⇔2x - \(6\sqrt{2x-1}\) + 3y- \(10\sqrt{3y-2}\)+ 4z-\(14\sqrt{4x-3}\) +77=0
⇔2x-1-\(6\sqrt{2x-1}\) +9 +3y-2-\(10\sqrt{3y-2}\)+25 +4z-3-\(14\sqrt{4x-3}\) +49=0
⇔\(\left(\sqrt{2x-1}-3\right)^2+\left(\sqrt{3y-2}-5\right)^2+\left(\sqrt{4y-3}-7\right)^2\) =0
Có: \(\left(\sqrt{2x-1}-3\right)^2\) ≥ 0 , ∀ x ;
\(\left(\sqrt{3y-2}-5\right)^2\) ≥ 0, ∀ y ;
\(\left(\sqrt{4z-3}-7\right)^2\) ≥ 0, ∀ z
⇒ \(\left(\sqrt{2x-1}-3\right)^2+\left(\sqrt{3y-2}-5\right)^2+\left(\sqrt{4y-3}-7\right)^2\) ≥0, ∀ x, y, z.
⇒\(\left\{{}\begin{matrix}\left(\sqrt{2x-1}-3\right)^2=0\\\left(\sqrt{3y-2}-5\right)^2=0\\\left(\sqrt{4z-3}-7\right)^2=0\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}\sqrt{2x-1}-3=0\\\sqrt{3y-2}-5=0\\\sqrt{4x-3}-7=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\sqrt{2x-1}=3\\\sqrt{3y-2}=5\\\sqrt{4z-3}=7\end{matrix}\right.\)⇔ \(\left\{{}\begin{matrix}2x-1=9\\3y-2=25\\4z-3=49\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=5\\y=9\\z=13\end{matrix}\right.\)
