Ta có : \(\frac{12x^2+12x+11}{4x^2+4x+3}=\frac{5y^2-10y+9}{y^2-2y+2}\)
\(\Leftrightarrow\frac{3\left(4x^2+4x+3\right)+2}{4x^2+4x+3}=\frac{5\left(y^2-2y+2\right)-1}{y^2-2y+2}\)
\(\Leftrightarrow3+\frac{2}{4x^2+4x+3}=5-\frac{1}{y^2-2y+2}\)
Do \(\frac{2}{4x^2+4x+3}=\frac{2}{\left(2x+1\right)^2+2}\le\frac{2}{2}=1\) \(\Rightarrow3+\frac{2}{4x^2+4x+3}\le4\left(1\right)\)
\(\frac{1}{y^2-2y+2}=\frac{1}{\left(y-1\right)^2+1}\le\frac{1}{1}=1\) \(\Rightarrow5-\frac{1}{y^2-2y+2}\ge5-1=4\left(2\right)\)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow VT=VP=4\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\y-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=1\end{matrix}\right.\)
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