Question

tìm giá trị lớn nhất của biểu thức:

k, D=\(\dfrac{2\left|x\right|+3}{3\left|x\right|-1}\)

n, N=2+\(\dfrac{12}{3\left|x+5\right|+4}\)

Answer 1

\(D=\dfrac{2\left|x\right|+3}{3\left|x\right|-1}\)

\(\left\{{}\begin{matrix}\left|x\right|\ge0\Rightarrow2\left|x\right|\ge0\Rightarrow2\left|x\right|+3\ge3\\\left|x\right|\ge0\Rightarrow3\left|x\right|\ge0\Rightarrow3\left|x\right|-1\ge-1\end{matrix}\right.\)

\(MAX_D\Rightarrow MIN_{3\left|x\right|-1}\)

\(3\left|x\right|-1\in Z^+\)

\(\Rightarrow3x-1=1\)

\(\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)

\(\Rightarrow MAX_D=\dfrac{2.\left|\dfrac{2}{3}\right|+3}{3.\left|\dfrac{2}{3}\right|-1}=\dfrac{\dfrac{13}{3}}{1}=\dfrac{13}{3}\)