\(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=0\)

\(\Leftrightarrow3x=40\)

hay \(x=\dfrac{40}{3}\)

\(x^3+3x^2+3x+9=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\x^2+3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-3\\x^2=-3\Rightarrow ktm\end{cases}}\)

Vậy x=-3

ktm là gì vậy bạn

ktm là ko thỏa mãn

vì \(x^2\ge0\forall x\)

mà: x^2 =-3 

=> x^2 =-3 ko thỏa mãn

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

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

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

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

\(x.\left(x-3\right)-3x+9=0\)

\(x.\left(x-3\right)-3.\left(x-3\right)=0\)

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

a,\(x^2-3x+2\left(x-3\right)=0.\)

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

\(\Leftrightarrow x^2+x-6=0\)

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

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

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

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

a,\(\orbr{\begin{cases}x=0\\x+3=0\end{cases}\left(=\right)\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)

b,\(\orbr{\begin{cases}x+2=0\\3x-9=0\end{cases}}\)(=)\(\orbr{\begin{cases}x=-2\\3x=9\end{cases}\left(=\right)\orbr{\begin{cases}x=-2\\x=3\end{cases}}}\)

c,\(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\)(=) \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Tìm x , biết : 

x. ( x + 3 ) = 0 

<=> x = 0 hoặc x + 3 = 0

<=> x = 0 hoặc x = -3 

Vậy x € { 0 ; 3  }

(  x + 2 ) . ( 3x - 9 ) = 0 

<=> x + 2 = 0 hoặc 3x - 9 = 0 

<=> x = -2 hoặc 3x = 9 

<=> x = -2 hoặc x = 3 

Vậy x € {  -2 ; 3 }

( x - 2 ) . ( x + 1 ) = 0 

<=> x - 2 = 0 hoặc x + 1 = 0 

<=> x = 2 hoặc x = -1 

Vậy x € { 2 ; -1 }

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

\(\text{Vậy x =0 hoặc x = -3}\)

\(\left(x+2\right).\left(3x-9\right)=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0-2\\3x=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=3\end{cases}}}\)

\(\text{Vậy x = -2 hoặc x = 3}\)

\(\left(x-2\right).\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

\(\text{Vậy x = 2 hoặc x = -1}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy x=3/2 hoặc x=-5

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)

b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)

c) \(5x\left(2x-3\right)-6x+9=0\)

\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\5x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)

a: Ta có: \(\left(2x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

b: Ta có: \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{3}\end{matrix}\right.\)

c: Ta có: \(5x\left(2x-3\right)-6x+9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

a, 5\(x\)(\(x^2\) - 9) = 0

    \(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)

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

Vậy \(x\) \(\in\) { -3; 0; 3}

b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0

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

    (\(x+3\))( 3 - \(x\)) = 0

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

Vậy \(x\) \(\in\){ -3; 3}

c, \(x^2\) - 9\(x\) - 10 = 0

   \(x^2\) + \(x\) - 10\(x\)  - 10 = 0

   \(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0

        (\(x+1\))(\(x-10\)) = 0

         \(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)

           \(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

Vậy \(x\) \(\in\){ -1; 10}

a) 5x(x2-9)=0

=> TH1 5x=0

      <=> x= 0

     TH2: 2x-9=0

       <=> 2x=9

       <=> x = \(\dfrac{9}{2}\)

b, 3(x+3) - x²- 3x = 0   

<=> 3x + 9 - x² -3x = 0

<=> - x² +9 = 0

<=> - x² = -9

 <=> x = 3

c, x² -9x -10 = 0

<=> x² -x + 10x -10 = 0

<=> x(x-1)+10(x-1)=0     

 <=> (x-1)(x+10)=0

=> TH1: x-1=0

       <=> x=1

      TH2: x +10=0

      <=>  x=-10

\(\left(\frac{2}{3}x-\frac{4}{9}\right).\left(\frac{1}{2}+\frac{-3}{7}\div x\right)=0\)

\(\left(\frac{2}{3}x-\frac{4}{9}\right).\left(\frac{1}{2}+\frac{-3}{8}.\frac{1}{x}\right)=0\)

\(\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}.x=0\end{cases}\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{4}{9}\\\frac{1}{x}=\frac{7}{6}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}}\)

\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=0\\ \Rightarrow\left(x^3-27\right)+x\left(4-x^2\right)=0\\ \Rightarrow x^3-27+4x-x^3=0\\ \Rightarrow4x-27=0\\ \Rightarrow4x=27\\ \Rightarrow x=\dfrac{27}{4}\)

\(b,\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\\ \Rightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\\ \Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Rightarrow12x+6=0\\ \Rightarrow12x=-6\\ \Rightarrow x=-\dfrac{1}{2}\)

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² = 4

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

<=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Vậy S = { 5 ; 1 }

b) x² - 9 = 0

<=> x² = 9

<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy S = { 3 ; -3 }

c) x( x - 2x ) - x² - 8 = 0

<=> x² - 2x² - x² - 8 = 0

<=> -2x² - 8 = 0

<=> -2x² = 8

<=> x² = -4 ( vô lí )

<=> x = \(\varnothing\)

Vậy S = { \(\varnothing\)}

d) 2x( x - 1 ) - 2x² + x - 5 = 0

<=> 2x² - 2x - 2x² + x - 5 = 0

<=> -x - 5 = 0

<=> -x = 5

<=> x = -5

Vậy S = { -5 }

e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0 

<=> x² - 3x - ( x² - x - 2 ) = 0

<=> x² - 3x - x² + x + 2 = 0

<=> - 2x + 2 = 0

<=> -2x = -2

<=> x = 1

Vậy S = { 1 }

f) x( 3x - 1 ) - 3x² - 7x = 0

<=> 3x² - x - 3x² - 7x = 0

<=> -8x = 0

<=> x = 0

Vậy S = { 0 } 

d) 2x(x - 1) - 2x² + x - 5 = 0

=> 2x² - 2x - 2x² + x - 5 = 0

=> -x = 5

=> x = -5

e) x(x - 3) - (x + 1)(x - 2) = 0

=> x² - 3x - (x² - x - 2) = 0 

=> x² - 3x - x² + x + 2 = 0

=> -2x = - 2

=> x = 1

f) x(3x - 1) - 3x² - 7x = 0

=> 3x² - x - 3x² - 7x = 0

=> -8x = 0

=> x = 0