a: \(\Leftrightarrow\left[{}\begin{matrix}2x-5=3-8x\\2x-5=8x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}10x=8\\-6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

1/ Ta có :

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

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

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

Để E đạt GTNN thì \(\left\{{}\begin{matrix}\left|4x-3\right|\\\left|5y+7,5\right|\end{matrix}\right.\) đạt GTNN

Dấu "=" xảy ra khi :

\(\left\{{}\begin{matrix}\left|4x-3\right|=0\\\left|5y+7,5\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-3=0\\5y+7,5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,75\\y=-1,5\end{matrix}\right.\)

Vậy GTNN của E bằng 17,5 khi \(\left\{{}\begin{matrix}x=0,75\\y=-1,5\end{matrix}\right.\)

Làm mẫu 1 phần :

a) \(|3x-1|+|x-1|=4\left(1\right)\)

Ta có: \(3x-1=0\Leftrightarrow x=\frac{1}{3}\)

             \(x-1=0\Leftrightarrow x=1\)

Lập bảng xét dấu :

+) Với \(x< \frac{1}{3}\Rightarrow\hept{\begin{cases}3x-1< 0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|3x-1|=1-3x\\|x-1|=1-x\end{cases}\left(2\right)}}\)

Thay (2) vào (1) ta được :

\(\left(1-3x\right)+\left(1-x\right)=4\)

\(2-4x=4\)

\(4x=-2\)

\(x=\frac{-1}{2}\)( chọn )

+) Với \(\frac{1}{3}\le x< 1\Rightarrow\hept{\begin{cases}3x-1>0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|3x-1|=3x-1\\|x-1|=1-x\end{cases}\left(3\right)}}\)

Thay (3) vào (1) ta được :
\(\left(3x-1\right)+\left(1-x\right)=4\)

\(2x=4\)

\(x=2\)( chọn )

+) Với \(x\ge1\Rightarrow\hept{\begin{cases}3x-1>0\\x-1>0\end{cases}\Rightarrow}\hept{\begin{cases}|3x-1|=3x-1\\|x-1|=x-1\end{cases}\left(4\right)}\)

Thay (4) vào (1) ta được :

\(\left(3x-1\right)+\left(x-1\right)=4\)

\(4x-2=4\)

\(4x=6\)

\(x=\frac{3}{2}\)( chọn )

Vậy \(x\in\left\{\frac{-1}{2};2;\frac{3}{2}\right\}\)

Ta có: \(|2x-3|-|3x-2|=0\)

\(\Leftrightarrow|2x-3|=|3x-2|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x-2\\2x-3=2-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-2+3\\2x+3x=2+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=-1\\5x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}\Leftrightarrow}x=1\)

a) \(\left|\left|x-1\right|-1\right|=2\Rightarrow\orbr{\begin{cases}\left|x-1\right|-1=2\\\left|x-1\right|-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}\left|x-1\right|=3\\\left|x-1\right|=-1\left(l\right)\end{cases}}\)

TH1: x - 1 = 3

         x      = 4

TH2: x - 1 = - 3

        x       = - 2 

b) Tương tự câu a.

c) \(\left|\left|2x-3\right|-x+1\right|=42-8\)

\(\left|\left|2x-3\right|-x+1\right|=34\)

TH1: \(\left|2x-3\right|-x+1=34\)

\(\left|2x-3\right|-x=33\)

Với \(x\ge\frac{3}{2}\), ta có \(2x-3-x=33\Rightarrow x=36\)  (tm)

Với \(x< \frac{3}{2}\), ta có \(3-2x-x+1=34\Rightarrow-3x=30\Rightarrow x=-10\left(tm\right)\)

TH2: \(\left|2x-3\right|-x+1=-34\)

\(\left|2x-3\right|-x=-35\)

Với \(x\ge\frac{3}{2}\), ta có \(2x-3-x=-35\Rightarrow x=-32\)  (l)

Với \(x< \frac{3}{2}\), ta có \(3-2x-x+1=-34\Rightarrow-3x=38\Rightarrow x=\frac{38}{3}\left(l\right)\)

d) Tương tự câu c.

\(\left|2x-3\right|-\left|3x+2\right|=0\)

\(\Leftrightarrow\left|2x-3\right|=\left|3x+2\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3x=2+3\\2x+3x=-2+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=5\\5x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)

                  Vậy \(x\in\left\{-5;\frac{1}{5}\right\}\)