A

A

A

a, -154+(x-9-18)=40

x-27=40-(-154)

x-27=194

x=194+27

x=221

b,\(\left|9-x\right|\)=64+(-7)

\(\left|9-x\right|\)=57

TH1:

9-x=57

x=9-57

x=-48

TH2:

9-x=-57

x=9-(-57)

x=66

a, -154+(x-9-18)=40

x-27=40-(-154)

x-27=194

x=194+27

x=221

b,=64+(-7)

=57

TH1:

9-x=57

x=9-57

x=-48

TH2:

9-x=-57

x=9-(-57)

x=66

a) \(x^2-4x=0\)

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

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

b) \(4x^2-9=0\)

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

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

d) \(x\left(2x+9\right)-4x-18=0\)

\(x\left(2x+9\right)-2\left(2x+9\right)=0\)

\(\left(2x+9\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

e) \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}}\)

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

\(x.\left(x-4\right)=0\)

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

\(4x^2-9=0\)

\(2^2x^2-9=0\)

\(\left(2x\right)^2-9=0\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

\(x\left(2x+9\right)-4x-18=0\)

\(x\left(2x+9\right)-\left(4x+18\right)=0\)

\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)

\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)

\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

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

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

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

\(\)

a)=> x² + 6x + 9 - x² + 4x = 39 => 10x = 30 => x = 3
b) => x(x - 9) + 2(x -9) = 0 => (x+2)(x-9) = 0 
+)th1: x + 2 = 0 => x = -2
+)th2: x - 9 =0 => x = 9 

\(a)\dfrac{-4}{9}-\dfrac{5}{9}:x=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{-8}{18}+\dfrac{-1}{18}\)

\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{-1}{2}\)

\(\Leftrightarrow x=\dfrac{5}{9}.\left(-2\right)\)

\(\Leftrightarrow x=\dfrac{-10}{9}\)

\(b)\left|x\right|=\dfrac{1}{2}\)

\(\Leftrightarrow x\in\left\{-\dfrac{1}{2};\dfrac{1}{2}\right\}\)

\(c)x\left(x-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{0;\dfrac{1}{3}\right\}\)

a) x=\(\dfrac{-10}{9}\)

b)x∈{\(\dfrac{1}{2}\); \(\dfrac{-1}{2}\)}

1,7-(18+x)=-15

<=>18+x=7+15

<=>18+x=22

<=>x=4

2,(x-17)-(-3)=0

<=>x-17+3=0

<=>x-17=-3

,<=>x=14

<=>

1, 

7 - ( 18 + x ) = - 15 

18 + x = 7 + 15 

18 + x = 22

x = 22 - 18 

x = 4 

2,

( x - 17 ) - ( - 3 ) = 0 

( x - 17 ) + 3 = 0 

x - 17 = - 3 

x = -3 + 17 

x = 14 

3, -18 - ( x - 6 ) = 0 

-18 = x- 6 

x = - 18 + 6 

x = - 12 

4,   29 - ( 10 + 29 ) = x - ( 27 - 9 )

-10 = x - 18 

x = - 10 + 18 

x = 8

( 3.x – 18).( x – 9) = 0

=> 3x - 18 = 0                                                    hoặc x - 9 = 0

      3x         = 0 + 18                                                          x = 0 + 9

      3x         =   18                                                               x = 9

        x         = 18 : 3

        x         = 6

Vậy x = 9 hoặc x = 6

Nếu a . b = 0 thì a hoặc b hoặc cả a và b phải là 0.

=> (3.x - 18) . (x - 9) = 0

Ta có:

* Nếu 3.x - 18 = 0 thì 3.x = 0 + 18 => 3.x = 18 => x = 18 : 3 => x = 6

* Nếu x - 9 = 0 thì x = 9

Vì cả 2 trường hợp trên x đều khác nhau nên không có trường hợp 3.x - 18 và x - 9 đều bằng 0.

( 3.x - 18 ) . ( x - 9 ) = 0 

( 3.x -18 )  . x          = 0 + 9 

( 3.x - 18 )  . x          = 9

 4.x - 18                   = 9

 4.x                          = 9 + 18  

 4.x                          = 27 

  x                            = 27 : 4

  x                            = 6,75

vậy x = 6,75

chúc bạn học tốt

\(\left(3x-18\right)\left(x-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-18=0\\x-9=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=18\\x=9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=9\end{cases}}\)

x=6 hoặc x=9 nha

( 3x - 18) . (9 - x) = 0

\(\Rightarrow\hept{\begin{cases}3x-18=0\\9-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=6\\x=9\end{cases}}}\)

( 3.x-18).(9-x)=0

<=>3x-18=0 hoặc 9-x=0

TH1: 3x-18=0

<=>3x=18

<=>x=6

Th2 :x-9=0

<=>x=9

vậy x=6 hoặc x=9

k mk nha mk mk bị âm điểm rùi

Ta có: ( 3.x - 18) . (9 - x) = 0

\(\Leftrightarrow\orbr{\begin{cases}3x-18=0\\9-x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0+18\\x=9-0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=18\\x=9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=9\end{cases}}\)