3

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

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

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

\(x^2-2018x=0\\\Leftrightarrow x\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2018=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2018\end{matrix}\right.\)

Vậy `x=0` hoặc `x=2018`

x.(x - 2018) = 0 

=> \(\left[{}\begin{matrix}x=0\\x-2018=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x=0+2018\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x=2018\end{matrix}\right.\)

Vậy x ϵ { 0 ; 2018 }

x² - 2x - 15 = 0

x² - 25 - 2x + 10 =0

( x² - 25) - ( 2x -10) =0

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

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

(x-5)(x+3)

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

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

kết luận x \(\in\) { -3; 5}

\(x^2-2018x=0\\ \Leftrightarrow x\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2108=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2018\end{matrix}\right.\)

Vậy `x=0` hoặc `x=2018`

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

Vậy `x=0` hoặc `x=-5/2`

\(x^2=100\)

\(\Leftrightarrow x=\pm\sqrt{100}=\pm10\)

x=10 hoặc x=-10

Ta có: \(x^2=100\)

nên \(x\in\left\{10;-10\right\}\)

Vậy: \(x\in\left\{10;-10\right\}\)

( x + 5 ) . ( x + 6 ) = 0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)

Vậy \(x=-5\) hoặc \(x=-6\)

8x - 9x -2x - 15 = 0

\(\Rightarrow8x-9x-2x=0+15\)

\(\Rightarrow-3x=15\)

\(\Rightarrow x=15:\left(-3\right)\)

\(\Rightarrow x=-5\)

a, \(\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)

Vậy ......

\(\left(x+5\right)\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)

\(8x-9x-2x-15=0\)

\(\Rightarrow8x-9x-2x=15\)

\(\Rightarrow-3x=15\)

\(\Rightarrow x=-5\)

\(8x^3+12x^2+6x-26=0\)

<=>  \(4x^3+6x^2+3x-13=0\)

<=>  \(4x^3-4x^2+10x^2-10x+13x-13=0\)

<=>  \(4x^2\left(x-1\right)+10x\left(x-1\right)+13\left(x-1\right)=0\)

<=>  \(\left(x-1\right)\left(4x^2+10x+13\right)=0\)

<=> \(x-1=0\)

<=>  \(x=1\)

Vậy...

Cam on ban nhieu nhe <3

Ta có :

 Với x chẵn => x = 2 => 2² + 117 = y²

  => 121 = y² => 11² = y² => y = 11 (thoả mãn)

Với x lẻ => x² cũng lẻ => x² + 117 chẵn và x > 2

=> y² chẵn => y = 2

Mà x < y => ko thoả mãn

Vậy x = 2 ; y = 11

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)