c) x^2 -x-20=0

\(\Leftrightarrow x^2-5x+4x-20=0\)

\(\Leftrightarrow\left(x^2+4x\right)-\left(5x+20\right)=0\)

\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=5\end{matrix}\right.\)

Vậy...

\(\left|x-3\right|+\left|x-5\right|=2x\)

\(\Leftrightarrow x-3+x-5=2x\)

\(\Leftrightarrow\left(x+x\right)-\left(3+5\right)=2x\)

\(\Leftrightarrow2x-8=2x\)

\(\Leftrightarrow2x-2x=8\)

\(\Leftrightarrow0x=8\) (vô lí)

\(\Leftrightarrow\) ko tìm dc x thỏa mãn

\(\left|x-3\right|+\left|x-5\right|=2x\)

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

\(\Rightarrow2x\ge0\)

\(\Rightarrow x-3+x-5=2x\)

\(\Rightarrow2x-8=2x\)

\(\Rightarrow x\in\varnothing\)

\(\left|x+2\right|+\left|2x-3\right|=5\)

+ Với \(x< -2\)Ta có \(-x-2-2x+3=5\)

\(-3x=5-3+2\)

\(-3x=4\)

\(x=-\frac{4}{3}\)( loại )

+Với \(-2\le x< \frac{3}{2}\)ta có \(x+2-2x+3=5\)

\(x-2x=5-3-2\)

\(x=0\)( nhận )

+ Với \(x\ge\frac{3}{2}\)ta có \(x+2+2x-3=5\)

\(3x=5+3-2\)

\(3x=6\)

\(x=2\)( nhận )

Vậy x=0 và x=2

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 \(\left|2x+3\right|=\left|11-2x\right|\)

\(\Rightarrow\left|2x+3\right|-\left|11-2x\right|=0\)

Ta thấy \(\left|2x+3\right|\ge0\); \(\left|11-2x\right|\ge0\)nên \(\left|2x+3\right|-\left|11-2x\right|\ge0\)

Mà theo bài ra \(\left|2x+3\right|-\left|11-2x\right|=0\)nên \(2x+3-11+2x=0\)

\(\Rightarrow\left(2x+2x\right)+3-11=0\)

\(\Rightarrow4x-8=0\Rightarrow4x=8\Rightarrow x=2\)

Vậy x = 2

Ta có:

|2x+3|=|11-2x|

=> 2x+3=11-2x hoặc 2x+3=-(11-2x)

Với 2x+3=11-2x                               Với 2x+3= -(11-2x)

=> 2x+2x = 11-3                                     2x+3= -11+2x

     4x      =8                                          2x-2x=-11-3

       x      =2                                             0   =-14(vô lí)

Vậy x=2