Vì:  |2\(x\) - 1| = |1 - 2\(x\)|

Nên:      |2\(x\) - 1| + |1 - 2\(x\)| = 8

  ⇒        |2\(x\) - 1| + |2\(x\) - 1| = 8

                         2.|2\(x\) - 1| = 8

                            |2\(x\) - 1| = 8:2

                           |2\(x\) - 1| = 4

                         \(\left[{}\begin{matrix}2x-1=-4\\2x-1=4\end{matrix}\right.\)

                          \(\left[{}\begin{matrix}2x=-4+1\\2x=4+1\end{matrix}\right.\)

                            \(\left[{}\begin{matrix}2x=-3\\2x=5\end{matrix}\right.\)

                             \(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\){- \(\dfrac{3}{2}\); \(\dfrac{5}{2}\)}

giúp mình với

a: ĐKXĐ: x<>2; x<>0

b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)

\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{x^3-2x^2-2x^2+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)

\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)

c: M>=-3

=>(x+1+6x)/2x>=0

=>(7x+1)/x>=0

=>x>0 hoặc x<=-1/7

Ta có: \(\left(2x-1\right)^3-\left(2x+1\right)\left(4x^2-2x+1\right)=-8\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3-1=-8\)

\(\Leftrightarrow-12x^2+6x+6=0\)

\(\Leftrightarrow2x^2-x-1=0\)

a=2; b=-1; c=-1

Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là: 

\(x_1=1;x_2=\dfrac{c}{a}=\dfrac{-1}{2}\)

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

/a/ là gì

Giá trị tuyệt đối à

\(+2x\ge1\Rightarrow|2x-1|+|1-2x|=2x-1+2x-1=2\left(2x-1\right)=8\Leftrightarrow2x-1=4\Leftrightarrow x=2,5\)

\(+2x\le1\Rightarrow|2x-1|+|1-2x|=1-2x+1-2x=2\left(1-2x\right)\Leftrightarrow1-2x=4\Leftrightarrow x>1\left(loại\right)\)

\(Vậy:x=2,5\)

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

\(\Rightarrow2x-2x=8-\left(1+1\right)\Leftrightarrow0=6\)(vô lí)

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

\(\Rightarrow2x-2x=\left(-8\right)-\left(1+1\right)\Leftrightarrow0=\left(-10\right)\)(vô lí)

Vậy ko có x thõa mãn

Ta có: |2x-1|=|1-2x|

=> |2x-1|+|1-2x|=8

=> |2x-1|+|2x-1|=8

=> 2|2x-1|=8

=> |2x-1|=4

=> \(\orbr{\begin{cases}2x-1=4\\2x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-3}{2}\end{cases}}}\)

Ta có:

|Ax|+|Bx| >_ |Ax+Bx|

=>|2x-1|+|1-2x| >_ |2x-1+1-2x|=0

=>0=8 (vô lí)

=>Không có giá trị của x thỏa mãn phương trình.

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

vì 2x-1 và 1-2x là 2 số đối => |2x-1| = |1-2x|

|2x-1| + |1-2x| = 8

=> 2|2x-1| = 8

=> |2x-1| = 4

\(\Rightarrow\begin{cases}2x-1=4\\2x-1=-4\end{cases}\\ \Rightarrow\begin{cases}2x=5\\2x=-3\end{cases}\\ \Rightarrow\begin{cases}x=2,5\\x=-1,5\end{cases}\)

vậy x = { 2,5;-1,5 }