[Toán 8] Toán nâng cao đại số 8

T

transformers123

$y=\dfrac{2x+7}{x^2+8}$

$\iff 8y=\dfrac{16x+56}{x^2+8}$

$\iff 8y=\dfrac{(x+4)^2}{x^2+8}-1$

$\iff 8y \ge -1$

$\iff y \ge \dfrac{-1}{8}$

Lại có: $y=\dfrac{2x+7}{x^2+8}$

$\iff y=\dfrac{-x^2+2x-1}{x^2+8}+1$

$\iff y=\dfrac{-(x-1)^2}{x^2+8}+1$

$\iff y \le 1$

Vậy $\dfrac{-1}{8} \le y \le 1$

Mà $y \in Z$ nên $\left[\begin{matrix}y=0\\y=1\end{matrix}\right. \Longrightarrow
\left[\begin{matrix}\dfrac{2x+7}{x^2+8}=0\\\dfrac{2x+7}
{x^2+8}=1\end{matrix}\right. $

Dến đây giải pt bình thường
 
You must log in or register to reply here.
Top Bottom