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

\(\Leftrightarrow-3x+5=0\)

hay \(x=\dfrac{5}{3}\)

b: Ta có: \(x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

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

a. x(x-2)+x-2=0

=> (x-2).(x+1)=0

=> x-2=0           hoặc x+1=0

=> x=2              hoặc x=-1

b. 5x(x-3)-x+3=0

=> 5x(x-3)-(x-3)=0

=> (x-3).(5x-1)=0

=> x-3=0         hoặc 5x-1=0

=> x=3            hoặc x=1/5

a) x = 2

b) x = 3

a) x(x - 2) + x - 2 = 0;

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

<=>(x-2)(x+1)=0

<=>x-2=0 hoặc x+1=0

<=>x=2 hoặc x=-1

b) 5x(x - 3) - x + 3 = 0

<=>5x.(x-3)-(x-3)=0

<=>(x-3)(5x-1)=0

<=>x-3=0 hoặc 5x-1=0

<=>x=3 hoặc x=1/5

x(x-2) + x-2= x(x-2) + (x-2).1=(x-2)(x+1)

à như thế này

 x (x-2) + 1( x - 2) = 0

 (x - 2)( x + 1) = 0 

a)  \(x\left(x-2\right)+x-2=0\)

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

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

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

Vậy...

b)  \(5x\left(x-3\right)-x+3=0\)

<=>  \(\left(x-3\right)\left(5x-1\right)=0\)

<=>  \(\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)

Vậy...

\(a,\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a,`

`(x - 2)(x - 3) =0`

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

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

`<=>`\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy, `S = {2; 3}`

`b,`

`x^2 - 5x = 0`

`<=> x(x - 5) = 0`

`<=>`\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=0+5\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

Vậy, `S = {0; 5}`

`c,`

`x^2 - 9 = 0`

`<=> x^2 = 0 + 9`

`<=> x^2 = 9`

`<=> x^2 = (+-3)^2`

`<=> x = +-3`

Vậy, `S = {3; -3}`

`d,`

`4x^2 - 25 = 0`

`<=> 4x^2 = 25`

`<=> x^2 = 25/4`

`<=> x^2 = (+-5/2)^2`

`<=> x = +-5/2`

Vậy,` S = {5/2; -5/2}.`

a: =>x-2=0 hoặc x-3=0

=>x=2 hoặc x=3

b: =>x(x-5)=0

=>x=0 hoặc x=5

c: =>(x-3)(x+3)=0

=>x=3 hoặc x=-3

d: =>(2x-5)(2x+5)=0

=>x=5/2 hoặc x=-5/2

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

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

Hoặc x - 2 = 0 => x = 2

Hoặc x + 1 = 0 => x = -1

Vậy x = -1; x = 2.

b) 5x(x - 3) - x + 3 = 0

5x(x - 3) - (x - 3) = 0

(x - 3)(5x - 1) = 0

Hoặc x - 3 = 0 => x = 3

Hoặc 5x - 1 = 0 => x = 1/5.

Vậy x = 1/5; x = 3.

Ta có:
a) \(x\left(x-2\right)+x-2=0\)
\(\Rightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+1=0\\x-2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-1\\x=2\end{cases}}\)

b) \(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\hept{\begin{cases}5x-1=0\\x-3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

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

    x ( x - 2 ) + ( x - 2 ) . 1 = 0

    ( x - 2 ) ( x + 1 ) = 0

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

Vậy x = 2 ; x = -1

b) 5x ( x - 3 ) - x + 3 = 0

5x ( x - 3 ) - ( x - 3 ) . 1 = 0

( x - 3 ) ( 5x - 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\5x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)

Vậy x = 3 ; x = 1/5

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

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

     ( x + 1 ) ( x - 2 ) = 0

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

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

Vậy \(x\in\left\{-1;2\right\}\)

b) 5x ( x - 3 ) - x + 3 = 0

 5x ( x - 3 ) - ( x - 3 ) = 0

      ( 5x - 1 ) ( x - 3 ) = 0

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{5};3\right\}\)