Giải phương trình `a, 4x^2 + 8x - 5 = 0 ` `b, 3x^2 - 2x - 1 = 0 ` `c, ( x^2 + x + 1 ) ( 5 - 3x ) = 0`

Answer

`a, 4x^2 + 8x - 5 = 0`

`<=> 4x^2 + 10x - 2x - 5 = 0`

`<=> (4x^2 + 10x) - (2x + 5) = 0`

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

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

`<=>` $\left[\begin{matrix} 2x - 1 = 0\\ 2x + 5 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 2x = 0 + 1\\ 2x = 0 - 5\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 2x = 1\\ 2x = - 5\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = 1 : 2\\ x = - 5 : 2\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = \dfrac{1}{2}\\ x = \dfrac{-5}{2}\end{matrix}\right.$

Vậy `S = {1/2 ; {-5}/2}`

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`b, 3x^2 - 2x - 1 = 0`

`<=> 3x^2 - 3x + x - 1 = 0`

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

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

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

`<=>` $\left[\begin{matrix} 3x + 1 = 0\\ x - 1 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 3x = 0 - 1\\ x = 0 + 1\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 3x = - 1\\ x = 1\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = - 1 : 3\\ x = 1\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = \dfrac{-1}{3}\\ x = 1\end{matrix}\right.$

Vậy `S = {{-1}/3 ; 1}`

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`c, (x^2 + x + 1) . (5 - 3x) = 0`

Trường hợp `1:`

`x^2 + x + 1 = 0`

`<=> x^2 + x + 1/4 + 3/4 = 0`

`<=> (x^2 + x + 1/4) + 3/4 = 0`

`<=> [x^2 + 2 . x . 1/2 + (1/2)^2] + 3/4 = 0`

`<=> (x + 1/2)^2 + 3/4 = 0`

Vì `(x + 1/2)^2 >= 0 AA x`

`=> (x + 1/2)^2 + 3/4 >= 0 AA x`

`=> x \in \phi`

Trường hợp `2:`

`5 - 3x = 0`

`<=> 3x = 5 - 0`

`<=> 3x = 5`

`<=> x = 5 : 3`

`<=> x = 5/3`

Vậy `S = {5/3}`