X - { [ -x + (x+3) ] } - [ (x+3) - (x-2)] = 0

X  -  {  -x + x + 3 } -  [ x +3 - x +2] = 0

X  - 3  - 5 = 0

x - 8 = 0

x = 8 

a) Ta có: \(\left(-12\right)-\left|13-x\right|=-21\)

\(\Leftrightarrow-\left|x-13\right|-12=-21\)

\(\Leftrightarrow-\left|x-13\right|=-9\)

\(\Leftrightarrow\left|x-13\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=9\\x-13=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=22\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{22;4\right\}\)

b) Ta có: \(8\le\left|x\right|< 9\)

\(\Leftrightarrow\left|x\right|=8\)

\(\Leftrightarrow x\in\left\{8;-8\right\}\)

Vậy: \(x\in\left\{8;-8\right\}\)

c) Ta có: \(x-\left(-25+x\right)=13-x\)

\(\Leftrightarrow x+25-x-13+x=0\)

\(\Leftrightarrow x+12=0\)

hay x=-12

Vậy: x=-12

d) Ta có: \(\left(15-30\right)+x=x-\left(27-\left|-8\right|\right)\)

\(\Leftrightarrow x-15=x-\left(27-8\right)\)

\(\Leftrightarrow x-15-x+19=0\)

\(\Leftrightarrow-4=0\)(vô lý)

Vậy: \(x\in\varnothing\)

\(8x-48+4x-12-14=-x+4\)

\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)

\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)

\(\Leftrightarrow2x=130\Leftrightarrow x=65\)

\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)

\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)

\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)

a: Ta có: \(2x\left(x-3\right)+x-3=0\)

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

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

b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)

\(\frac{-6}{30}=\frac{x}{-20}\)

nhân chéo  \(x\cdot30=\left(-6\right)\cdot\left(-20\right)\)

=>\(30x=120\)

\(x=4\)

\(\frac{-6}{30}=\frac{3}{y}\)

nhân chéo => \(-6x=90\)

\(x=-15\)

\(\frac{-6}{30}=\frac{z}{5}\)

nhân chéo => \(30z=-30\)

\(z=-1\)

x/-20 = -6/30 

=> 30x = 120 

<=> x = 4 

3/y = -6/30 

=> -6y = 90 

<=> y = -15 

z/5 = -6/30 

=> -6z = 150 

<=> z = - 25 

x=4,y=-15,z=1

1. áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\frac{x+2}{3}=\frac{y-7}{5}=\frac{x+y-5}{3+5}=\frac{16}{8}=2\Rightarrow\hept{\begin{cases}x+2=6\\y-7=10\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=17\end{cases}}}\)

2. áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-y+2}{2-3}=\frac{-10+7}{-1}=3\Rightarrow\hept{\begin{cases}x+5=6\\y-2=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=11\end{cases}}\)

a) (x - 5)(x - 3) + 2(x - 5) = 0

(x - 5)(x - 3 + 2) = 0

(x - 5)(x - 1) = 0

x - 5 = 0 hoặc x - 1 = 0

*) x - 5 = 0

x = 5

*) x - 1 = 0

x = 1

Vậy x = 1; x = 5

b) (x - 2)(x² + 2x + 4) - (x + 2)(x² - 2x + 4) = 2(x + 2)

x³ - 8 - x³ - 8 = 2x + 4

2x = -8 - 8 - 4

2x = -20

x = -20 : 2

x = -10

a)

\(\left(x-5\right)\left(x-3\right)+2\left(x-5\right)=0\)

\(\left(x-5\right)\left(x-3+2\right)=0\)

\(\left(x-5\right)\left(x-1\right)=0\)

\(x-5=0\) hoặc \(x-1=0\)

+) \(x-5=0\\ \Rightarrow x=5\)

+) \(x-1=0\\ \Rightarrow x=1\)

Vậy \(x=1\) hoặc \(x=5\)

b) \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2-2x+4\right)=2\left(x+2\right)\)

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

\(2x=-8-8-4\)

\(2x=-20\)

 \(x=-20:2\)

 \(x=-10\)

Vậy \(x=-10\)