x - {[-x+(x+3)]} + [(x+3)-(x-2)] = 0
<=> x - (-x+x+3) + (x+3-x+2) = 0
<=> x - 3 + 5 = x + 2 = 0
<=> x = -2

\(A=\dfrac{x^3-2x^2-15x}{x-5}=\dfrac{x\left(x^2-2x-15\right)}{x-5}=\dfrac{x\left(x+3\right)\left(x-5\right)}{x-5}=x\left(x+3\right)\)

\(A=x^2+3x=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)

\(A_{min}=-\dfrac{9}{4}\)

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

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

    3(x+3)-x(x+3)=0

     (x+3)(3-x)     =0

      x+3 =0 hoặc 3-x=0   =>x={-3;3}

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

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

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

vậy_____

Vì (x+1).(x-2)=0

suy ra 1 trong 2 so =0

tu giai not

a, vì (x+1).(x-2)=0 nên x+1=0  hoặc x-2=0

nếu x+1=0               nếu x-2=0

      x    =0-1                  x    =0+2

      x    =-1                    x    =2

Có ai giúp tôi với😭😭😭😭😭

b) \(x^3-6x^2+9x=0\)

\(\Leftrightarrow x.\left(x^2-6x+9\right)=0\)

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

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

Vậy \(x=0\)hoặc \(x=3\)

a. ( x - 1 )³ + 1 + 3x ( x - 4 ) = 0

<=> x³ - 3x² + 3x - 1 + 1 + 3x² - 12x = 0

<=> x³ - 9x = 0

<=> x ( x² - 9 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

b. x³ - 6x² + 9x = 0

<=> x ( x² - 6x + 9 ) = 0

<=> x ( x - 3 )² = 0

<=> \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

a) ( x - 1 )³ + 1 + 3x( x - 4 ) = 0

⇔ x³ - 3x² + 3x - 1 - 1 + 3x² - 12x = 0

⇔ x³ - 9x = 0

⇔ x( x² - 9 ) = 0

⇔ \(\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

b) x³ - 6x² + 9x = 0

⇔ x( x² - 6x + 9 ) = 0

⇔ x( x - 3 )² = 0

⇔ \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

a, => x^3 < 0 ; x-3 > 0 hoặc x^3 > 0 ; x-3 < 0

=> 0 < x < 3

b, => x^4.(2x-8) < 0

=> x^4.(x-4) < 0

Vì x^4 >= 0

=> x-4 < 0

=> x  < 4

c, Vì x-1 < x+12

=> x-1 < 0 ; x+12 >0

=> -12 < x < 1

d, => x-12 > 0 ; x-1 > 0 hoặc x-12 < 0 ; x-1 < 0

=> x  >12 hoặc x < 1

Tk mk nha

Thank you so much