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

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`

3

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

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

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

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}

Đề:............

<=> - (1 - 2018x) + 2019x.(1 - 2018x) = 0

<=> (1 - 2018x).[(-1) + 2019x] = 0

Xét 2 trường hợp, ta có:

TH₁: 1 - 2018x = 0          TH₂: -1 + 2019x = 0

<=> 2018x = 1                 <=> 2019x = 1

<=> x = 1/2018                <=> x = 1/2019

Vậy x = 1/2018; 1/2019

\(2018x-1+2019x\left(1-2018x\right)=0\)

\(-\left(1-2018x\right)+2019x\left(1-2018x\right)=0\)

\(\left(1-2018x\right)\left(-1+2019x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}1-2018x=0\\-1+2019x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2018}\\x=\frac{1}{2019}\end{cases}}}\)

\(x^2=100\)

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

x=10 hoặc x=-10

Ta có: \(x^2=100\)

nên \(x\in\left\{10;-10\right\}\)

Vậy: \(x\in\left\{10;-10\right\}\)

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

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

x = 2017 nha

Ta có : x(x-2017)-2018x+2017.2018=0

=>x(x-2017)-2018(x-2017)=0

=>(x-2017)(x-2018)=0

=>x=2017;2018.