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a, \(A=\left|x-1\right|+\left|x+1\right|+\left|x-2\right|+\left|x-3\right|\ge\left|1-x+x+1\right|+\left|2-x+x-3\right|=3\)
Dấu ''='' xảy ra khi \(\left(1-x\right)\left(x+1\right)\ge0;\left(2-x\right)\left(x-3\right)\ge0\Leftrightarrow-1\le x\le1;2\le x\le3\Leftrightarrow-1\le x\le3\)
Vậy GTNN của A bằng 3 tại -1 =< x =< 3
b, \(B=\left|x+1\right|+\left|x-1\right|+\left|2x-5\right|\ge\left|x+1+x-1\right|+\left|2x-5\right|\)
\(=\left|2x\right|+\left|2x-5\right|=\left|2x\right|+\left|5-2x\right|\ge\left|2x+5-2x\right|=5\)
Dấu ''='' xảy ra khi \(\left(x+1\right)\left(x-1\right)\ge0;2x\left(5-2x\right)\ge0\Leftrightarrow;0\le x\le\frac{5}{2}\)
Vậy GTNN của B bằng 5 tại 0 =< x =< 5/2
Ta có : |5x + 1| + |3 - 2x| \(\ge\left|5x+1+3-2x\right|=\left|4+3x\right|\)
Dấu "=" xảy ra <=> \(\left(5x+1\right)\left(3-2x\right)\ge0\)
Xét các trường hợp
TH1 : \(\hept{\begin{cases}5x+1\ge0\\3-2x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge-0,2\\x\le1,5\end{cases}}\Rightarrow-0,2\le x\le1,5\)
TH2 :\(\hept{\begin{cases}5x+1\le0\\3-2x\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le-0,2\\x\ge1,5\end{cases}}\Rightarrow x\in\varnothing\)
Vậy \(-0,2\le x\le1,5\)là giá trị cần tìm
Ta có : \(\left|x+\frac{2}{3}\right|=\frac{3}{5}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{3}{5}\\x+\frac{2}{3}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}-\frac{2}{3}\\x=-\frac{3}{5}-\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{15}\\x=-\frac{19}{15}\end{cases}}\)
/x/+2/3=3/5 hoặc /x/+2/3=-3/5
x=3/5-2/3 x=-3/5-2/3
x=-1/15 x=-19/15
/x/-2,8=1/5 hoặc /x/-2,8=-1/5
x=1/5+2,8 x=-1/5+2,8
x=3 x=13/5
/x/+1/2+3=0
x+7/2=0
x=0-7/2
x=-7/2
/2x/-3/8=0
2x=0+3/8
2x=3/8
x=3/8:2
x=3/16
\(\left|x-3,2\right|+\left|\dfrac{2x-1}{5}\right|=x+3\) (1)
TH1: \(\left\{{}\begin{matrix}x>3,2\Rightarrow\left|x-3,2\right|=x-3,2\\x>\dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{2x-1}{5}\end{matrix}\right.\)
\(\left(1\right)\Rightarrow x-3,2+\dfrac{2x-1}{5}=x+3\)
\(\Rightarrow5x-16+2x-1=5x+15\Rightarrow2x=32\Leftrightarrow x=16\left(tm\right)\)
TH2: \(\left\{{}\begin{matrix}x>3,2\Rightarrow\left|x-3,2\right|=x-3,2\\x< \dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{1-2x}{5}\end{matrix}\right.\)
\((1)\)\(\Rightarrow x-3,2+\dfrac{1-2x}{5}=x+3\Rightarrow5x-16+1-2x=5x+15\)
\(\Rightarrow-2x=0\Rightarrow x=0\left(l\right)\)
TH3: \(\left\{{}\begin{matrix}x< 3,2\Rightarrow\left|x-3,2\right|=3,2-x\\x>\dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{2x-1}{5}\end{matrix}\right.\)
\(\left(1\right)\Rightarrow3,2-x+\dfrac{2x-1}{5}=x+3\)
\(\Rightarrow16-5x+2x-1=5x+15\Rightarrow8x=0\Leftrightarrow x=0\left(l\right)\)
TH4: \(\left\{{}\begin{matrix}x< 3,2\Rightarrow\left|x-3,2\right|=3,2-x\\x< \dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{1-2x}{5}\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3,2-x+\dfrac{1-2x}{5}=x+3\)
\(\Rightarrow16-5x+1-2x=5x+15\Rightarrow12x=2\Rightarrow c=\dfrac{1}{6}\left(tm\right)\)
Vậy \(x=\left\{16;\dfrac{1}{6}\right\}\)