\(Q=\dfrac{3-4x}{x^2+1}=\dfrac{4x^2+4-4x^2-4x-1}{x^2+1}=\dfrac{4\left(x^2+1\right)-\left(2x+1\right)^2}{x^2+1}\)\(=4-\dfrac{\left(2x+1\right)^2}{x^2+1}\). Do \(\left(2x+1\right)^2\ge0;x^2+1>0\Rightarrow Q\le4\)
Vậy Max Q = 4 \(\Leftrightarrow x=-\dfrac{1}{2}\)
Cách khác:
\(Q=\dfrac{3-4x}{x^2+1}\)
\(\Leftrightarrow Qx^2+Q-3+4x=0\)(*)
+)Xét Q=0=>\(x=\dfrac{3}{4}\)
+)Xét Q\(\ne0\)
Để pt(*) có nghiệm thì \(\Delta=16-4Q\left(Q-3\right)\ge0\)
\(\Leftrightarrow4-Q^2+3Q\ge0\)
\(\Leftrightarrow\left(4-Q\right)\left(Q+1\right)\ge0\)
\(\Leftrightarrow-1\le Q\le4\)
\(\Rightarrow MAXQ=4\Leftrightarrow x=-\dfrac{1}{2}\)
