Question

Tìm giá trị lớn nhất của biểu thức:

A= 4 - |5x - 2| - |3g + 12|

Answer 1

3g p ko

Answer 2

\(\left\{{}\begin{matrix}-\left|5x-2\right|\le0\\-\left|3g+12\right|\le0\end{matrix}\right.\Rightarrow-\left|5x-2\right|-\left|3g+12\right|\le0\)

\(\Rightarrow A=4-\left|5x-2\right|-\left|3g+12\right|\le4\)

Dấu " = " khi \(\left\{{}\begin{matrix}\left|5x-2\right|=0\\\left|3g+12\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\g=-4\end{matrix}\right.\)

Vậy \(MAX_A=4\) khi \(\left\{{}\begin{matrix}x=\dfrac{2}{5}\\g=-4\end{matrix}\right.\)

Answer 3

\(A=4-\left|5x-2\right|-\left|3g+12\right|\)

\(\left|5x-2\right|\ge0;\left|3g+12\right|\ge0\)

\(A_{MAX}\Rightarrow\left|5x-2\right|_{MIN};\left|3g+12\right|_{MIN}\)

\(\left|5x-2\right|_{MIN}=0;\left|3g+12\right|_{MIN}=0\)

\(\Rightarrow A_{MAX}=4-0-0=4\)