Lời giải:

a. $22-(-x)=12$

$22+x=12$

$x=12-22=-10$

b. $x(x+2)=0$

$\Rightarrow x=0$ hoặc $x+2=0$

$\Rightarrow x=0$ hoặc $x=-2$

c. $(x+1)(x+9)=0$

$\Rightarrow x+1=0$ hoặc $x+9=0$

$\Rightarrow x=-1$ hoặc $x=-9$

d.

$x^2+3x=0$

$\Rightarrow x(x+3)=0$

$\Rightarrow x=0$ hoặc $x+3=0$

$\Rightarrow x=0$ hoặc $x=-3$

a) 22 - (-x) = 12

x = 12 - 22

x = -10

b) x.(x + 2) = 0

x = 0 hoặc x + 2 = 0

*) x + 2 = 0

x = 0 - 2

x = -2

Vậy x = -2; x = 0

c) (x + 1)(x + 9) = 0

x + 1 = 0 hoặc x + 9 = 0

*) x + 1 =.0

x = 0 - 1

x = -1

*) x + 9 = 0

x = 0 - 9

x = -9

Vậy x = -9; x = -1

d) x² + 3x = 0

x(x + 3) = 0

x = 0 hoặc x + 3 = 0

*) x + 3 = 0

x = 0 - 3

x = -3

Vậy x = -3; x = 0

9. Tìm x biết 

a x- ( 12- 15)= -8

<=> x - 12 + 15 = -8

<=> x + 3 = -8

<=> x = -11

Vậy x = -11

b. (x- 29 )- ( 17- 38)= -9

<=> x - 29 - 17 + 38 = -9

<=> x - 8 = -9

<=> x = -1

Vậy x = -1

a) x-12+15=-8

=>x=-8-15+12

=>x=-11

Vậy x=-11

b)x-29+21=-9

=>x=-9-21+29

=>x=-1

Vậy x=-1

a) Ta có: x-(12-15)=-8

\(\Leftrightarrow x-12+15=-8\)

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

hay x=-8-3=-11

Vậy: x=-11

b) Ta có: \(\left(x-29\right)-\left(17-38\right)=-9\)

\(\Leftrightarrow x-29-17+38+9=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: x=-1

a,x=-6 x=-12

b,x=-33

c,x= 2015 x= 2016

a)\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

   \(\frac{1}{3}:2x=-5-\frac{1}{4}\)

   \(\frac{1}{3}:2x=-\frac{21}{3}\)

   \(2x=\frac{1}{3}:\left(\frac{-21}{3}\right)\)

   \(2x=-\frac{1}{21}\)

   \(x=\frac{-1}{42}\)

b)\(\left(3x-\frac{1}{4}\right).\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{12}\\x=-\frac{1}{2}\end{array}\right.\)

c)\(\left(2x-5\right).\left(\frac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

   \(\Rightarrow\left[\begin{array}{nghiempt}2x-5=0\\\frac{3}{2}x+9=0\\0,3x-12=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=5\\\frac{3}{2}x=-9\\0,3x=12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-6\\x=40\end{array}\right.\)

a) 1/4 + 1/3 : 2x = -5

=> 1/3 : 2x = -5 - 1/4

=> 1/3 : 2x = -21/4

=> 2x = 1/3 : (-21/4) = -4/63

=> x = -4/63 : 2 = -2/63

\(x\left(x-3\right)-12+4x=0\)

\(\Leftrightarrow x^2-3x-12+4x=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

x = 2 hoặc = 6

Cách làm:

x² - 8x + 12 = 0

x² - 6x - 2x + 12 = 0

( x² - 6x ) - ( 6x - 12 ) = 0

x . ( x - 2 ) - 6 . ( x - 2 ) = 0

( x - 2 ) . ( x - 6 ) = 0

\(\Rightarrow\hept{\begin{cases}x-2=0\\x-6=0\end{cases}}\hept{\begin{cases}x=2\\x=6\end{cases}}\)

x²-8x+12=0

<=> x²-6x-2x+12=0

<=> (x²-2x)-(6x-12)=0

<=> x.(x-2)-6.(x-2)=0

<=> (x-2)(x-6)=0

<=>\(\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

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

\(=x\left(x-2\right)-6\left(x-2\right)\)

\(=\left(x-2\right)\left(x-6\right)\)

X+4/12 = 3/-9

X         =3/-9-1/3

X         =-2/3

Ta có:

\(\left(x+5\right)\left(3x-12\right)=\left(x+5\right)3\left(x-4\right)=3.\left[\left(x+5\right)\left(x-4\right)\right]\)

Để \(3.\left[\left(x+5\right)\left(x-4\right)\right]<0\) thì x+5 và x-4 trái dấu.

Mà x+5>x-4 

\(\Rightarrow x+5>0\) và \(x-4<0\)

\(\Rightarrow x>-5\) và \(x<4\)

x là số nguyên ta có \(x\in\left\{-4;-3;-2;-1;1;2;3\right\}\)

Vậy \(x\in\left\{-4;-3;-2;-1;1;2;3\right\}\)

X=1 , X=2 , X=3