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\(\text{Ta có:}A=\frac{2010x+2680}{x^2+1}=\frac{-335x^2+335x^2-335+2010x+2680+335}{x^2+1}.\)
\(=\frac{-335\left(x^2+1\right)+335\left(x^2+6x+9\right)}{x^2+1}=-335+\frac{335\left(x+3\right)^2}{x^2+1}\ge-335\)
\(\text{Vậy GTNN của A=-335. Dấu bằng xảy ra khi và chỉ khi }x+3=0\Leftrightarrow x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=\frac{2010x+2680}{x^2+1}\)
\(\Leftrightarrow Ax^2-2010x+A-2680=0\)
\(\Delta=\left(-2010\right)^2-4A\left(A-2680\right)\)
\(=-4\left(A-3015\right)\left(A+335\right)\)
Có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow\left(A-3015\right)\left(A+335\right)\le0\)
\(\Leftrightarrow\hept{\begin{cases}A\le3015\\A\ge-335\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:A=\(\frac{335x^2+2010x+3015-\left(335x^2+335\right)}{x^2+1}\)
= \(\frac{335\left(x^2+6x+9\right)}{x^2+1}-\frac{335\left(x^2+1\right)}{x^2+1}\)
=\(\frac{335\left(x+3\right)^2}{x^2+1}-335\)
Ta có: (x+3)2>= 0
=>335(x+3)2>=0
Mà x2+1>0
=>\(\frac{335\left(x+3\right)^2}{x^2+1}\ge0\)
=>\(\frac{335\left(x+3\right)^2}{x^2+1}-335\ge-335\)
=>A>= -335
Dấu "=" xảy ra <=> (x+3)2=0
<=> x+3=0
<=> x=-3
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{2010x+2680}{x^2+1}=\frac{-335x^2-335}{x^2+1}+\frac{335x^2+2010x+3015}{x^2+1}=\)
\(=-335+\frac{335\left(x^2+6x+9\right)}{x^2+1}=-335+\frac{335\left(x+3\right)^2}{x^2+1}\)
Vì \(\left(x+3\right)^2_{min}=0\Leftrightarrow335\left(x+3\right)^2_{min}=0\Leftrightarrow\frac{335\left(x+3\right)^2}{x^2+1}_{min}=0\Leftrightarrow-335+\frac{335\left(x+3\right)^2}{x^2+1}_{min}=-335\)
Vậy \(\frac{2010x+2680}{x^2+1}_{min}=-335\) tại x=-3.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2010x+2690}{x^2+1}=-335+\frac{335\left(x+3\right)^2}{x^2+1}\ge-335\)
Vậy giá trị nỏ nhất của A là : -335 khi x= -3
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)