Đặt C(x)=0

\(\Leftrightarrow-2x\left(2x-3\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow-4x^2+6x-2x+2=0\)

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

\(\Leftrightarrow4x^2-4x-2=0\)

\(\Leftrightarrow\left(2x-1\right)^2=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}\\2x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{3}+1\\2x=-\sqrt{3}+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+1}{2}\\x=\dfrac{-\sqrt{3}+1}{2}\end{matrix}\right.\)

Đặt Q(x)=0

\(\Leftrightarrow2\left(x-3\right)-\left(x-1\right)=0\)

\(\Leftrightarrow2x-6-x+1=0\)

\(\Leftrightarrow x=5\)

\(\left(x+3\right)\left(2x-1\right)-\left(x-4\right)\left(2x+1\right)=10\)

\(\Leftrightarrow2x^2-x+6x-3-\left(2x^2+x-8x-4\right)=10\)

\(\Leftrightarrow2x^2-x+6x-3-2x^2-x+8x+4=10\)

\(\Leftrightarrow12x+1=10\)

\(\Leftrightarrow12x=9\)

\(\Leftrightarrow x=\frac{9}{12}\)

\(\Rightarrow x=\frac{3}{4}\)

Vậy   \(x=\frac{3}{4}\)

Lời giải:
Ta thấy: 

$2015x=|2x+1|+|2x+2|+...+|2x+1007|\geq 0$

$\Rightarrow x\geq 0$.

Khi đó:

$|2x+1|=2x+1, |2x+2|=2x+2, |2x+3|=2x+3,...., |2x+1007|=2x+1007$.

Phương trình trở thành:

$(2x+1)+(2x+2)+....+(2x+1007)=2015x$

$\Rightarrow \underbrace{(2x+2x+...+2x)}_{1007}+(1+2+3+...+1007)=2015x$

$\Rightarrow 2x.1007+1007.1008:2=2015x$

$\Rightarrow 2014x+507528=2015x$

$\Rightarrow x=507528$

Vế trái có 101 số hạng

VT=(2x+1+2x+101).101:2

=(4x+102).101:2

=>(4x+102).101:2=5757

(4x+102).101=5757.2

(4x+102).101=11514

4x+102=11514:101

4x+102=114

4x=114-102

4x=12

x=12:4

x=3

Vậy x thuộc {3}

ve trai co 101 so hang 

vt:(2x+1+2x101).101:2

=(4x+102).101:2

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

Trường hợp 1: x<3/2

Pt sẽ là 3-2x+2-x=2x+1

=>-3x+5=2x+1

=>-5x=-4

hay x=4/5(nhận)

Trường hợp 2: 3/2<=x<2

Pt sẽ là 2x-3+2-x=2x+1

=>2x+1=x-1

=>x=-2(loại)

Trường hợp 3: x>=2

Pt sẽ là x-2+2x-3=2x+1

=>3x-5=2x+1

hay x=6(nhận)

\(\left|x-\dfrac{1}{2}\right|\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)

\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\\\left|2x-\dfrac{3}{4}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow2x-\dfrac{3}{4}\ge0\)

\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)

\(\Rightarrow\left|x-\dfrac{1}{2}\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=1\Rightarrow x=\dfrac{3}{2}\\x-\dfrac{1}{2}=-1\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)

\(2x-\dfrac{3}{4}\ge0\Rightarrow2x\ge\dfrac{3}{4}\Rightarrow x\ge\dfrac{3}{2}\)

Vậy xảy ra khi:

\(x=\dfrac{3}{2}\)