\(x^2+2=6\)

\(x^2=6-2\)

\(x^2=4\)

\(x=-2\) hoặc \(x=2\)

Vậy \(S=\left\{-2;2\right\}\)

\(x^2\) + 2 = 6

\(x^2\) = 6 - 2

\(x^2\) = 4

Vì \(x^2\) = 4 nên x = 2 hoặc -2

Vậy x={2;-2}.

Vì sao hiện nay nguời tả ưu tiên dùng xe điện

ilovevietnam

100,999

100+0,99999= 100,99999

6\(x\) - 18 > 4\(x\) - 6

6\(x\) - 4\(x\) > - 6 + 18

   2\(x\) > 12

    \(x>12:2\)

    \(x\) > 6

Vậy \(x\) > 6 

`(2x - 5)(2x + 1) = (2x - 5)(x + 4)`

`(2x - 5)(2x + 1) - (2x - 5)(x +4) = 0`

`(2x - 5)[(2x + 1) - (x + 4)]=0`

`(2x - 5)(2x + 1 - x - 4) = 0`

`(2x - 5)(x - 3) = 0`

\(\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\\ \left[{}\begin{matrix}2x=5\\x=3\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)

(2\(x-5\)).(2\(x+1\)) = (2\(x-5\)).(\(x+4\))

(2\(x-5\))(2\(x+1\)) - (\(2x-5\)).(\(x+4\)) = 0

(2\(x-5\))[2\(x+1\) - \(x-4\)] = 0

 (2\(x-5\)).[(2\(x-x\)) - (4 - 1)] = 0

  (2\(x\) - 5).[\(x\) - 3] = 0

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

    \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\) \(\in\) {\(\dfrac{5}{2}\); 3} 

a; (\(\sqrt{45}\) - \(\sqrt{125}\) + \(\sqrt{20}\)) : \(\sqrt{5}\)

   =  (\(\sqrt{9.5}\) - \(\sqrt{25.5}\) + \(\sqrt{4.5}\)):\(\sqrt{5}\)

   =  (3\(\sqrt{5}\) - 5\(\sqrt{5}\) + 2\(\sqrt{5}\)): \(\sqrt{5}\)

     = (- 2\(\sqrt{5}\) + 2\(\sqrt{5}\)) : \(\sqrt{5}\)

     = 0 : \(\sqrt{5}\)

     = 0 

cuu t