\(x^2=100\)

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

x=10 hoặc x=-10

Ta có: \(x^2=100\)

nên \(x\in\left\{10;-10\right\}\)

Vậy: \(x\in\left\{10;-10\right\}\)

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`

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^2-2y^2+2/3x^2y^3+B=2x^2+y^2+2/3x^2y^3`

`=>B=2x^2+y^2+2/3x^2y^3-x^2+2y^2-2/3x^2y^3`

`=>B=(2x^2-x^2)+(y^2+2y^2)+(2/3x^2y^3-2/3x^2y^3)`

`=>B=x^2+3y^2`

Thay `x=1 ; y=[-1]/3` vào `B` có:

   `B=1^2+3.([-1]/3)^2=1+3 . 1/9=1+1/3=4/3`

`x^2 - 2y^2 + 2/3x^2y^3 + B = 2x^2 + y^2 + 2/3x^2y^3`

`=> B  = 2x^2 + y^2 + 2/3x^2y^3` `- (x^2 - 2y^2 + 2/3x^2y^3)`

         `= 2x^2 + y^2 + 2/3x^2y^3 - x^2 + 2y^2 - 2/3x^2y^3`

         `= ( 2x^2 - x^2 ) + ( y^2 + 2y^2 ) + ( 2/3x^2y^3 - 2/3x^2y^3 )`

         `= x^2 + 3y^2`

Thay `x=1 ; y=-1/3` vào `B` ta có `:`

`B = 1^2 + 3 . ( -1/3 )^2`

   `= 1 + 1/3`

   `= 4/3` 

8^2=64 =>x=2

7^3=343 =>x=7

x = 2

và x = 7

Ta có: 8^x = 64

=> x = 2

Ta có: x³ = 343

=> x =7