1) \(|5x-3|=|7-x|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)

Vậy...

2) \(2.|3x-1|-3x=7\)

\(\Leftrightarrow2.|3x-1|=7+3x\)

\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)

Vậy...

5) \(\text{1,6 - |x + 1,5| = 0}\)

\(|x + 1,5| = 1,6-0\)

\(\text{|x + 1,5| = 1,6}\)

\( |x | = 1,6-1,5\)

\(|x|=0,1\)

a, +) Xét \(x\ge2,5\) có:
\(x-1,5+x-2,5=0\)

\(\Leftrightarrow2x=4\Leftrightarrow x=2\) ( không t/m )

+) Xét \(1,5\le x< 2,5\) có:
\(x-1,5+2,5-x=0\)

\(\Leftrightarrow1=0\) ( ko t/m )

+) Xét x < 1,5 có:

\(1,5-x+2,5-x=0\)

\(\Leftrightarrow2x=4\Leftrightarrow x=2\) ( ko t/m )

Vậy không có giá trị x thỏa mãn

b, \(\left\{{}\begin{matrix}\left|x+3\right|\ge0\\\left|x+1\right|\ge0\end{matrix}\right.\Leftrightarrow\left|x+3\right|+\left|x+1\right|\ge0\)

\(\Leftrightarrow3x\ge0\Leftrightarrow x\ge0\)

\(\Leftrightarrow x+3+x+1=3x\)

\(\Leftrightarrow x=4\)

Vậy x = 4

c, \(\left|x-7\right|=1-2x\)

+) Xét \(x\ge7\) có:
\(x-7=1-2x\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)( ko t/m )

+) Xét x < 7 có:

\(7-x=1-2x\Leftrightarrow x=-6\) ( t/m )

Vậy x = -6

\(\left|x-1,5\right|+\left|2,5-x\right|=0\)

\(\left|x-1,5\right|\ge0\) \(\left|2,5-x\right|\ge0\)

Dấu "=" xảy ra khi

\(\left|x-1,5\right|=0\Rightarrow x-1,5=0\Rightarrow x=1,5\)

\(\left|2,5-x\right|=0\Rightarrow2,5-x=0\Rightarrow x=2,5\)

\(\left|x+3\right|+\left|x+1\right|=3x\)

\(\left|x+3\right|\ge0;\left|x+1\right|\ge0\)

\(\Leftrightarrow\left|x+3\right|+\left|x+1\right|\ge0\)

\(\Rightarrow x+3+x+1=3x\)

\(2x+4=3x\)

\(x=4\)

a: =>|x-1/2|=2x+1

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left(2x+1\right)^2-\left(x-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left(2x+1-x+\dfrac{1}{2}\right)\left(2x+1+x-\dfrac{1}{2}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left(x+\dfrac{3}{2}\right)\left(3x-\dfrac{1}{2}\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

b: =>\(\left\{{}\begin{matrix}x-1.3=0\\2y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1.3\\y=\dfrac{1}{2}\end{matrix}\right.\)

a . ( x - 1/2 ) - 2 x = 1 

=> x - 1/2 = 1 hoặc 2x =0

=> x = 3/2 hoặc x = 0

b .( x -1/3 ) + ( 2y -1 ) = 0

=> x - 1/3 = 0 hoặc 2y - 1 = 0

=> x = 1/3 hoặc 2y = 1 

=> x = 1/3 hoặc y = 1/2

c. ( x - 1,5 ) + ( y - 2,5 ) + ( x + y + z ) nhỏ hơn hoặc bằng 0

=> x - 1,5 = 0  hoặc y - 2,5 = 0 hoặc x + y + z = 0

=> x= 1,5 hoặc y= 2,5 hoặc x + y +z = 0

=> x = 1,5 hoặc y = 2,5 hoặc 1,5 + 2,5 + z = 0 

=> x = 1,5 hoặc y = 2,5 hoặc z = 4 , - 4

a: \(\left(x+5\right)^2-\left(x-5\right)^2-2x+1=0\)

=>\(x^2+10x+25-\left(x^2-10x+25\right)-2x+1=0\)

=>\(x^2+8x+26-x^2+10x-25=0\)

=>18x+1=0

=>\(x=-\dfrac{1}{18}\)

b: \(\left(2x-7\right)^2-\left(x+3\right)^2=3x^2+6\)

=>\(4x^2-28x+49-\left(x^2+6x+9\right)-3x^2-6=0\)

=>\(x^2-28x+43-x^2-6x-9=0\)

=>34-34x=0

=>34x=34

=>x=1

c: \(\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\)

=>\(9x^2+12x+4-9\left(x^2-25\right)-225+5x=0\)

=>\(9x^2+17x+4-225-9x^2+225=0\)

=>17x+4=0

=>x=-4/17

\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

a) làm mẫu cho cả phần b lun

 \(|2x-5|+|2,5-x|=0\left(1\right)\)

Ta có: \(2x-5=0\Leftrightarrow x=\frac{5}{2}\)

          \(2,5-x=0\Leftrightarrow x=2,5=\frac{5}{2}\)

Lập bảng xét dấu :

+) Với \(x< \frac{5}{2}\Rightarrow\hept{\begin{cases}2x-5< 0\\2,5-x< 0\end{cases}\Rightarrow}\hept{\begin{cases}|2x-5|=5-2x\\|2,5-x|=x-2,5\end{cases}}\left(2\right)\)

Thay (2) vào (1) ta được :

\(5-2x+x-2,5=0\)

\(-x+\frac{5}{2}=0\)

\(x=\frac{5}{2}\)( loại ) 

+) Với \(x\ge\frac{5}{2}\Rightarrow\hept{\begin{cases}2x-5\ge0\\2,5-x\ge0\end{cases}\Rightarrow}\hept{\begin{cases}|2x-5|=2x-5\\|2,5-x|=2,5-x\end{cases}}\left(3\right)\)

Thay (3) vào (1) ta được :

\(2x-5+2,5-x=0\)

\(x-\frac{5}{2}=0\)

\(x=\frac{5}{2}\)( chọn )

Vậy \(x=\frac{5}{2}\)

a) |2x - 5| + |2,5 - x| = 0

2x - 5 = 0 hoặc 2,5 - x = 0

2x = 0 + 5         -x = 0 - 2,5

2x = 5               -x = -2,5

x = 2,5               x = 2,5

=> x = 2,5

b) |x - 1,5| + |x + 3| = 0

x - 1,5 = 0 hoặc x + 3 = 0

x = 0 + 1,5         x = 0 - 3

x = 1,5               x = -3

=> x = 1,5 hoặc x = -3

c) (5x - 2)² = 1

(5x - 2)² = 1²

5x - 2 = 1; -1

5x - 2 = 1 hoặc 5x - 2 = -1

5x = 1 + 2         5x = -1 + 2

5x = 3               5x = 1

x = 3/5              x = 1/5

=> x = 3/5 hoặc x = 1/5

d) (4x - 1)³ + 7 = -20

(4x - 1)³ = -20 - 7

(4x - 1)³ = -27

(4x - 1)³ = (-3)³

4x - 1 = -3

4x = -3 + 1

4x = -2

x = -2/4 = -1/2

a)Ta có : \(\left|2x-5\right|\ge0\forall x\)

\(\left|2,5-x\right|\ge0\)

\(\Rightarrow\left|2x-5\right|+\left|2,5-x\right|=0\Leftrightarrow\hept{\begin{cases}\left|2x-5\right|=0\\\left|2,5-x\right|=0\end{cases}}\)

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

\(\Rightarrow\hept{\begin{cases}2x=5\\x=2,5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=2,5\\x=2,5\end{cases}}\)

\(\Rightarrow x=2,5\)

b) Ta có : \(\left|x-1,5\right|\ge0\forall x\)

\(\left|x+3\right|\ge0\forall x\) 

\(\Rightarrow\left|x-1,5\right|+\left|x+3\right|=0\Leftrightarrow\hept{\begin{cases}\left|x-1,5\right|=0\\\left|x+3\right|=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-1,5=0\\x+3=0\end{cases}}\)

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

\(\Rightarrow x\in\varnothing\)

c) \(\left(5x-2\right)^2=1\)

\(\Rightarrow\left(5x-2\right)^2=1^2\)

\(\Rightarrow5x-2=\pm1\)

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

\(\Rightarrow\orbr{\begin{cases}5x=3\\5x=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0,6\\x=0,2\end{cases}}\)

d) \(\left(4x-1\right)^3+7=-20\)

\(\Rightarrow\left(4x-1\right)^3=-27\)

\(\Rightarrow\left(4x-1\right)^3=\left(-3\right)^3\)

\(\Rightarrow4x-1=-3\)

\(\Rightarrow4x=-2\)

\(\Rightarrow x=-0,5\)