a) | 2x - 1 | = 1- 3x

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

\(\orbr{\begin{cases}2x-3x=1+1\\2x-1=-1+3x\end{cases}}\)

\(\orbr{\begin{cases}-x=2\\2x+3x=-1+1\end{cases}}\)

\(\orbr{\begin{cases}x=-2\\5x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=0\end{cases}}\)

b) | 1 - 2x | = x + 1 

\(\orbr{\begin{cases}1-2x=x+1\\1-2x=-\left(x+1\right)\end{cases}}\)

\(\orbr{\begin{cases}-2x-x=1-1\\-2x+x=-1-1\end{cases}}\)

\(\orbr{\begin{cases}-3x=0\\-x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

tương tự

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

a)              \(x^2-5x+4=0\)

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

\(\Leftrightarrow\)\(x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=4\end{cases}}\)

Vậy tổng các giá trị nguyên của x thỏa mãn là:

                \(1+4=5\)

a, \(\left|x+2\right|-\left|x+7\right|=0\Rightarrow\left|x+2\right|=\left|x+7\right|\Rightarrow\orbr{\begin{cases}x+2=x+7\\x+2=-x-7\end{cases}\Rightarrow\orbr{\begin{cases}0=5\left(loại\right)\\2x=-9\end{cases}\Rightarrow}x=\frac{-9}{2}}\)

b, - Nếu \(2x-1\ge0\Rightarrow x\ge\frac{1}{2}\), ta có: 2x - 1 = 2x - 1 => 2x = 2x (thỏa mãn với mọi x)

- Nếu 2x - 1 < 0 => \(x< \frac{1}{2}\), ta có: 2x - 1 = 1 - 2x => 4x = 2 => x = \(\frac{1}{2}\) (không thỏa mãn điều kiện)

Vậy \(x\ge\frac{1}{2}\)

c,d tương tự b

e, tương tự a

a: Khi x>0 thì A=3x-3x+2=2

Khi x<0 thì A=-3x-3x+2=-6x+2

b: B=4-x-x+5=9-2x

c: TH1: 5/4<x<5/2

A=5-2x-3x+7=12-5x

TH2: x>=5/2

A=2x-5-3x+7=-x+2

d: D=3-5x+|5x-3|

TH1: x>=3/5

D=3-5x+5x-3=0

TH2: x<3/5

D=3-5x+3-5x=6-10x

a, Áp dụng t/c dtsbn:

\(5x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{y-x}{5-7}=\dfrac{2}{-2}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-7\\y=-5\end{matrix}\right.\)

b, Áp dụng t/c dtsbn:

\(\dfrac{x}{y}=\dfrac{7}{2}\Rightarrow\dfrac{x}{7}=\dfrac{y}{2}=\dfrac{x+y}{7+2}=\dfrac{-27}{9}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-21\\y=-6\end{matrix}\right.\)

c, \(\dfrac{x}{32}=\dfrac{2}{x}\Rightarrow x^2=2\cdot32=64\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

d, \(\left|x+\dfrac{1}{3}\right|-2=\dfrac{1}{2}\Rightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{5}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{5}{2}\\x+\dfrac{1}{3}=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=-\dfrac{17}{6}\end{matrix}\right.\)