a,x.(x+7)=0

suy ra x=o hoặc x+7=0

vs x+7=0

         x=0+7

         x=7

vậy x=0 hoặc x=7                                

b(2+2x)(7-x)=0

suy ra 2+2x=0 hoặc 7-x=0

vs2+2x=0        vs7-x=0

2x        =0-2            x=0+7

2x        =(-2)            x=7

           x=(-2);2

           x=-1

vậy x=-1 hoặc x=7

d(x^2-9)(3x+15)=0

suy ra x^2-9=0 hoặc 3x+15=0

vsx^2-9=0         vs 3x+15=0

    x^2   =0+9          3x     =0-15

    x^2   =9               3x    =-15

    x^2   =3^2                 x=(-15):3

suy ra x=3 hoặc x=-3    x=-5

vậy x=3 x=-3 hoặc x=-5

e,(4x-8)(x^2+1)=0

suy ra4x-8=0 hoặc x^2+1=0

vs  4x-8=0         vs x^2+1=0

     4x    =0+8          x^2    =0-1

     4x    =8              x^2    =-1

        x   =8:4           x^2     =-1^2 hoặc 1^2

        x    =2                suy ra x=-1 hoặc x=1

vậy x=2, x=-1 hoặc x=1        

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

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

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a) \(15-5\left|x+4\right|=-12-3\)

\(\Leftrightarrow5\left|x+4\right|=30\)

\(\Leftrightarrow\left|x+4\right|=6\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)

b) \(\left(4x-8\right)\left(7-x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-8=0\\7-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)

c) \(\left(x^2-36\right)\left(x^2+5\right)=0\Rightarrow\left(x-6\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

d) \(-3\left(x+7\right)-11=2\left(x+5\right)\)

\(\Leftrightarrow-3x-32=2x+10\)

\(\Leftrightarrow5x=-42\Rightarrow x=-\frac{42}{5}\)

a) $(x-3)^2-(x+2)(x-2)=-5$

$\Rightarrow x^2-2\cdot x\cdot3+3^2-(x^2-2^2)=-5$

$\Rightarrow x^2-6x+9-(x^2-4)=-5$

$\Rightarrow x^2-6x+9-x^2+4=-5$

$\Rightarrow-6x+13=-5$

$\Rightarrow-6x=-18$

$\Rightarrow x=3$

b) $x^3-2x^2-4x+8=0$

$\Rightarrow(x^3-2x^2)-(4x-8)=0$

$\Rightarrow x^2(x-2)-4(x-2)=0$

$\Rightarrow (x^2-4)(x-2)=0$

$\Rightarrow (x^2-2^2)(x-2)=0$

$\Rightarrow (x-2)(x+2)(x-2)=0$

$\Rightarrow (x-2)^2(x+2)=0$

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

$\text{#}Toru$

a, x : (-2) = 9

    x          = 9 . (-2)

    x          = -18

b, 4x + (-8) = 24

    4x           = 24 - (-8)

    4x           = 32

      x           = 32 : 4

      x           = 8

c, (3 - x)  . (x + 7) = 0

TH1: 3 - x = 0

              x = 3

TH2: x + 7 = 0

         x       = -7

a, x: (-2) = 9

x             = 9 . (-2)

x             = -18

b,

4x + (-8)  = 24

4x - 8 = 24

4x = 24 + 8

4x = 32

x = 32 /4

x= 8

c,

(3-x).(x+7) = 0

TH1 :

3-x = 0 

x = 3- 0

x = 3

TH2 :

x+7 = 0

x = 0-7

x= -7 

vậy \(x = {3, -7}\)

a) x : (-2) = 9

x = 9 . (-2)

x = -18

b) 4x + (-8) = 24

4x - 8 = 24

4x = 24 + 8

4x = 32

x = 32 : 4

x = 8

c) ( 3 - x ) . ( x + 7 ) = 0

3 - x = 0

x = 3 - 0

x = 3

hoặc

x + 7 = 0

x = 0 - 7

x = -7

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