help me !!!

\(\)bài nào có MIN or MAX thì mk làm,mk ko làm thì có nghĩa là ko có nha

\(D=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)

\(\left\{{}\begin{matrix}\left|4x-3\right|\ge0\\\left|5y+7,5\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|\ge0\)

\(\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|4x-3\right|=0\Rightarrow4x=3\Rightarrow x=\dfrac{3}{4}\\\left|5y+7,5\right|=0\Rightarrow5y=-7,5\Rightarrow y=-1,5\end{matrix}\right.\)

\(\Rightarrow MIN_D=17,5\) khi \(x=\dfrac{3}{4};y=-1,5\)

\(E=4-\left|5x-2\right|-\left|3y+12\right|\)

\(\left\{{}\begin{matrix}\left|5x-2\right|\ge0\\\left|3y+12\right|\ge0\end{matrix}\right.\)

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

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|5x-2\right|=0\Rightarrow5x=2\Rightarrow x=\dfrac{2}{5}\\\left|3y+12\right|=0\Rightarrow3y=-12\Rightarrow y=-4\end{matrix}\right.\)

\(\Rightarrow MAX_E=4\) khi \(x=\dfrac{2}{5};y=-4\)

Bài 1:

a, \(A=3,7+\left|4,3-x\right|\ge3,7\)

Dấu " = " khi \(\left|4,3-x\right|=0\Rightarrow x=4,3\)

Vậy \(MIN_A=3,7\) khi x = 4,3

b, \(B=\left|3x+\dfrac{41}{5}\right|-14,2\ge-14,2\)

Dấu " = " khi \(\left|3x+\dfrac{41}{5}\right|=0\Rightarrow x=\dfrac{-41}{15}\)

Vậy \(MIN_B=-14,2\) khi \(x=\dfrac{-41}{15}\)

c, \(C=\left|4x-3y\right|+\left|5y+7,5\right|\ge17,5\)

( do \(\left|4x-3y\right|+\left|5y+7,5\right|\ge0\) )

Dấu " = " khi \(\left\{{}\begin{matrix}\left|4x-3y\right|=0\\\left|5y+7,5\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{8}\\y=-1,5\end{matrix}\right.\)

Vậy \(MIN_C=17,5\) khi \(\left\{{}\begin{matrix}x=\dfrac{-9}{8}\\y=-1,5\end{matrix}\right.\)

Bài 2:

a, \(A=5,5-\left|2x-1,5\right|\le5,5\)

Dấu " = " khi \(\left|2x-1,5\right|=0\Rightarrow x=0,75\)

Vậy \(MIN_A=5,5\) khi x = 0,75

b, c tương tự

a. \(2x\left(x-5\right)-x\left(2x+3\right)=26\Rightarrow2x^2-10x-2x^2-3x=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b. \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\frac{5}{2}\)

\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3=\frac{5}{2}\)\(\Rightarrow2y=\frac{7}{2}\Rightarrow y=\frac{7}{4}\)

c. \(2x^2+3\left(x+1\right)\left(x-1\right)=5x^2+5x\Rightarrow5x^2-3=5x^2+5x\)

\(\Rightarrow x=-\frac{3}{5}\)

cảm ơn bạn nhiều nhé 

kb vs mình đi 

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

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}5x-2=0\\3y+12=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)

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

a: -4|x-2|<=0

=>-4|x-2|+10<=10

Dấu = xảy ra khi x=2

b: |x+2|>=0

=>-|x+2|<=0

=>E<=0
Dấu = xảy ra khi x=-2