Cho hàm số $y=\dfrac{2x+4}{x^2+4x+3}$. Phương trình $y''=0$ có nghiệm là
 | $x=-4$ |
 | $x=-2$ |
 | $x=0$ |
 | $x=2$ |
Chọn phương án B.
Ta có $y=\dfrac{2x+4}{x^2+4x+3}=\dfrac{2\left(x+2\right)}{\left(x+2\right)^2-1}$.
y'&=\left(\dfrac{2\left(x+2\right)}{\left(x+2\right)^2-1}\right)^{\prime}\\
&=\dfrac{2\left(x+2\right)^2-2-2\left(x+2\right).2\left(x+2\right)}{\left[\left(x+2\right)^2-1\right]^2}\\
&=\dfrac{-2\left(x+2\right)^2-2}{\left[\left(x+2\right)^2-1\right]^2}.
\end{aligned}$
$\begin{aligned}
y''&=\left(\dfrac{-2\left(x+2\right)^2-2}{\left[\left(x+2\right)^2-1\right]^2}\right)^{\prime}\\
&=\dfrac{-4\left(x+2\right)\left[\left(x+2\right)^2-1\right]^2-\left[-2\left(x+2\right)^2-2\right].2\left[\left(x+2\right)^2-1\right]2\left(x+2\right)}{\left[\left(x+2\right)^2-1\right]^4}\\
&=\dfrac{4\left(x+2\right)\left[\left(x+2\right)^2-1\right]\left[-\left(x+2\right)^2+1+2\left(x+2\right)^2+2\right]}{\left[\left(x+2\right)^2-1\right]^4}\\
&=\dfrac{4\left(x+2\right)\left[\left(x+2\right)^2+3\right]}{\left[\left(x+2\right)^2-1\right]^3}.
\end{aligned}$
Ta có $y''=0\Leftrightarrow\dfrac{4\left(x+2\right)\left[\left(x+2\right)^2+3\right]}{\left[\left(x+2\right)^2-1\right]^3}=0$.
Điều kiện $\left(x+2\right)^2-1\ne0$.
Khi đó $y''=0\Leftrightarrow x+2=0\Leftrightarrow x=-2$.