1 giải các phương trình sau.

l -) (2x-1)(x+2)+(x+2)(3-x): (- x) = 0

c) 2x * (x ^ 2 + 2) = (x - 3)(x ^ 2 + 2)

d x ^ 2 - 2x - 3 = 0

Đáp án:

$c) 2x.(x^2+2)=(x-3).(x^2+2)$

$⇔ 2x.(x^2+2)-(x-3).(x^2+2)=0$

$⇔ (x^2+2).(2x-x+3)=0$

 $⇔ (x^2+2).(x+3)=0$

$⇔$ \(\left[ \begin{array}{l}x^2+2=0\\x+3=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x^2=-2 \text{(Vô lí)}\\x=-3\end{array} \right.\) 

$\text{Vậy S={3}}$

$d) x^2-2x-3=0$

$⇔ x²+x-3x-3=0$

$⇔ x.(x-3)+1.(x-3)=0$

$⇔ (x+1).(x-3)=0$

$⇔$ \(\left[ \begin{array}{l}x+1=0\\x-3=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=-1\\x=3\end{array} \right.\) 

$\text{Vậy S={-1;3}}$

Đáp án 

`c) - 3 `

`d) - 1 ; 3 `

Giải thích các bước giải:

Câu `1` chắc sai đề. 

 `c) 2x ( x^2 + 2 ) = ( x - 3 ) ( x^2 + 2 ) `

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

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

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

`<=>`\(\left[ \begin{array}{l}x^2+2=0\\x+3=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x^2=-2( vô lý ) \\x=-3\end{array} \right.\) 

`=> x = - 3`

Vậy `S = { - 3 } `

`d) x^2 - 2x - 3 = 0 `

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

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

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

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

`<=> `\(\left[ \begin{array}{l}x+1=0\\x-3=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-1\\x=3\end{array} \right.\) 

Vậy `S = { - 1 ; 3 } `