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

\(x^2-2x-xy+2y=\left(x^2-xy\right)-2\left(x-y\right)=x\left(x-y\right)-2\left(x-y\right)=\left(x-y\right)\left(x-2\right)\)

5

ko có?

\(x^2=-25\)

Vì \(x^2\ge0\forall x\)

Mà \(x^2=-25\) (vô lí)

Vậy: \(x\in\varnothing\)

x(x-2)-5x+10=0

x(x-2)-5(x-2)=0

(x-5)(x-2)=0

x-5=0 hoặc x-2=0

x=5 hoặc x=2

<=> x^2-2x-5x+10=0

<=>x^2-7x+10=0

<=>(x-5).(x-2)=0 ( dùng cách bấm nghiệm bằng máy tính mode-5-3 , rồi nhập hệ số ) 

<=>x-5=0 hoặc x-2=0

<=>x=5      <=>x=2

Vậy PT có nghiệm là x=5 ; x=2

x(x-2) -5x +10 =0

x(x-2) -5x =10

x = 10:5

x=2

`|x+3|+10-5x=0`

`<=>|x+3|=5x-10(x>=2)`

`+)x+3=5x-10`

`<=>4x=13`

`<=>x=13/4(tm)`

`+)x-3=10-5x`

`<=>6x=13`

`<=>x=13/6(tm)`

Vậy `S={13/4,13/6}`

\(\left|x+3\right|+10-5x=0\)

\(\Leftrightarrow\left|x+3\right|=5x-10\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-10\ge0\\\left[{}\begin{matrix}x+3=5x-10\\x+3=10-5x\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=\dfrac{13}{4}\left(N\right)\\x=\dfrac{7}{6}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)

Giải:

\(\left|x+3\right|+10-5x=0\) 

\(\Rightarrow\left|x+3\right|=5x-10\) 

\(\Rightarrow\left[{}\begin{matrix}5x-10=x+3\\5x-10=x-3\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{4}\\x=\dfrac{7}{4}\end{matrix}\right.\) 

Vậy \(x\in\left\{\dfrac{13}{4};\dfrac{7}{4}\right\}\) 

Chúc bạn học tốt!

`|x+3|+10-5x=0`

`<=>|x+3|=5x-10(x>=2)`

`+)x+3=5x-10`

`<=>4x=13`

`<=>x=13/4(tm)`

`+)x-3=10-5x`

`<=>6x=13`

`<=>x=13/6(tm)`

Vậy `S={13/4,13/6}`

TH1: `x+3>=0 <=> x>=-3`

`x+3+10-5x=0`

`-4x=-13`

`x=13/4` (TM)

TH2: `x+3<0 <=> x<-3`

`-x-3+10-5x=0`

`-6x=-7`

`x=7/6` (L)

Vậy `x=13/4`

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

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

(5x - 10) . (6x + 12) = 0 

=>(5x - 10) = 0 hoặc (6x + 12) = 0 

 * TH1: (5x - 10) = 0           * TH2: (6x + 12) = 0 

             =>x = 2                             => x = -2

                     Vậy x=2 hoặc x = -2

a: Ta có: \(x^2+3x-10=0\)

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

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

b: Ta có: \(x^2-5x-6=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

3x( x - 2 ) + 10 - 5x = 0

<=> 3x( x - 2 ) - ( 5x - 10 ) = 0

<=> 3x( x - 2 ) - 5( x - 2 ) = 0

<=> ( x - 2 )( 3x - 5 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\3x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{5}{3}\end{cases}}\)

a) 2x² - 3x - 2 = 0.

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

<=> 2x + 1 = 0 hoặc x - 2 = 0

<=> x = -1/2 hoặc x = 2

b) 3x² - 7x - 10 = 0.

<=> (x + 1)(3x - 10) = 0

<=> x = -1 hoặc x = 10/3

c) 2x² - 5x + 3 = 0.

<=> (x - 1)(2x - 3) = 0

<=> x = 1 hoặc x = 3/2

2x( x + 1 ) - 5x - 10 = 0

<=> 2x² + 2x - 5x - 10 = 0

<=> 2x² - 3x - 10 = 0

<=> 2( x² - 3/2x + 9/16 ) - 89/8 = 0

<=> 2( x - 3/4 )² = 89/8

<=> ( x - 3/4 )² = 89/16

<=> \(\left(x-\frac{3}{4}\right)^2=\left(\pm\sqrt{\frac{89}{16}}\right)^2=\left(\pm\frac{\sqrt{89}}{4}\right)^2\)

<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{\sqrt{89}}{4}\\x-\frac{3}{4}=-\frac{\sqrt{89}}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{89}}{4}\\x=\frac{3-\sqrt{89}}{4}\end{cases}}\)