tìm x a) 2x(x-5)-12(5-x)=0 b) (5x-1)^2 - (2x+3)^2 =0

`a) 2x(x - 5) - 12(5 - x) = 0`

 `<=> 2x(x - 5) + 12(x - 5) = 0`

`<=> (x - 5)(2x + 12) = 0`

`<=>` $\left[\begin{matrix} x - 5 = 0\\ 2x + 12 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = 0 + 5\\ 2(x + 6) = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = 5\\ x + 6 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = 5\\ x = 0 - 6\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = 5\\ x = - 6\end{matrix}\right.$

Vậy `x in {5 ; -6}`

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`b) (5x - 1)^2 - (2x + 3)^2 = 0`

`<=> [(5x - 1) - (2x + 3)][(5x - 1) + (2x + 3) = 0`

`<=> (5x - 1 - 2x - 3)(5x - 1 + 2x + 3) = 0`

`<=> (3x - 4)(7x + 2) = 0`

`<=>` $\left[\begin{matrix} 3x - 4 = 0\\ 7x + 2 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 3x = 0 + 4\\ 7x = 0 - 2\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 3x =  4\\ 7x = - 2\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x =  \dfrac{4}{3}\\ x = \dfrac{-2}{7}\end{matrix}\right.$

Vậy `x in {4/3 ; -2/7}`

`text{#maingoctranthi}`