\(\sqrt[]{3+x}+\sqrt[]{6-x}=\sqrt[]{4x^2-12x+27}\left(đkxđ:-3\le x\le6\right)\)
\(\Leftrightarrow\left(\sqrt[]{3+x}+\sqrt[]{6-x}\right)^2=4x^2-12x+27\left(1\right)\)
Áp dụng Bđt Bunhiacopxki ta có :
\(VP=\left(1.\sqrt[]{3+x}+1.\sqrt[]{6-x}\right)^2\le\left(1^2+1^2\right)\left(3+x+6-x\right)=2.9=18\)
\(VP=4x^2-12x+27\)
\(\Leftrightarrow VP=4\left(x^2-3x+\dfrac{9}{4}\right)-9+27\)
\(\Leftrightarrow VP=4\left(x-\dfrac{3}{2}\right)^2+18\ge18,\forall x\in R\)
\(\left(1\right)\Leftrightarrow\left(\sqrt[]{3+x}+\sqrt[]{6-x}\right)^2=4x^2-12x+27=18\)
Dấu "=" xảy ra khi và chỉ khi :
\(\left\{{}\begin{matrix}\sqrt[]{3+x}=\sqrt[]{6-x}\\x-\dfrac{3}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3+x=6-x\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=3\\x=\dfrac{3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{3}{2}\left(tmđkxđ\right)\)
Vậy nghiệm của phương trình đã cho là \(x=\dfrac{3}{2}\)
