Mình làm lại nhé !
D= 4 - |5x - 2| - |3y + 12|
Ta có : 5x - 2 > 0 => -|5x -2|<0
3x + 12 >0 => -|3x + 12 |< 0
=> 4 - |5x -2 | - |3y + 12 |>4 hay D >4
\(\left[{}\begin{matrix}5x-2=0\\3y+12=0\end{matrix}\right.\)
Sau đó tính ra nhé !
Chúc bạn học tốt !
D= 4 - |5x - 2| - |3y + 12|
Ta có : 5x - 2 > 0 => -|5x -2|<0
3x + 12 >0 => -|3x + 12 |< 0
=> 4 - |5x -2 | - |3y + 12 |>4 hay D >4
=>\(\left[{}\begin{matrix}5x-2=0\\3y+12=0\end{matrix}\right.\)
=>
\(A=-\left|x\right|\le0\)
Dấu "=" xảy ra khi:
\(x=0\)
\(B=4-\left|5x-2\right|\le4\)
Dấu "=" xảy ra khi:
\(\left|5x-2\right|=0\Rightarrow x=\dfrac{2}{5}\)
\(C=\left|x-3\right|-\left|5-x\right|=\left|x-3\right|-\left|x-5\right|\)
\(C\le\left|x-3-x-5\right|=8\)
Dấu "=" xảy ra khi:
\(3\le x\le5\)
\(D=4-\left|5x-2\right|-\left|3y+12\right|\le4\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|5x-2\right|=0\\\left|3y+12\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)
\(E=5-3\left(2x-1\right)\le5\)
Dấu "=" xảy ra khi:
\(2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
