tìm x biết : a)x.(2x-3)-2.(3-2x)=0 b)(x+1/2)^2-(x+1/2).(x+6)=8 c) x^2.(xy+1)+2y-x-3xy

Đáp án:

 

Giải thích các bước giải:

$ a) x . ( 2x - 3 ) - 2 . ( 3 - 2x ) = 0 $

$ ⇔ x . ( 2x - 3 ) + 2 ( 2x - 3 ) = 0 $

$ ⇔ ( 2x - 3 ) ( x + 2 ) = 0 $  

$⇒\left[\begin{matrix} 2x - 3 = 0\\ x + 2 = 0\end{matrix}\right.$

$⇒\left[\begin{matrix} 2x = 3\\ x = - 2\end{matrix}\right. $

$\left[\begin{matrix} x = \frac{3}{2} \\ x=-2\end{matrix}\right.$

Vậy $ x ∈ { \frac{3}{2} ; - 2 } $

$ b)(x+\frac{1}{2})^{2} - ( x + \frac{1}{2}) . ( x + 6 ) = 8 $

$ ⇔ ( x+\frac{1}{2}) . ( x+\frac{1}{2}) - 1 . ( x+\frac{1}{2}) . ( x + 6 ) = 8 $

$ ⇔ ( x+\frac{1}{2}) . ( x+\frac{1}{2} - 1 ) . ( x + 6 ) = 8 $

$⇒ \left[\begin{matrix} x+\frac{1}{2}=8\\ x+\frac{1}{2} - 1=8\\ x+6=8 \end{matrix}\right.$

$⇒ \left[\begin{matrix} x=8 - \frac{1}{2}\\ x+\frac{1}{2} = 9\\ x=14 \end{matrix}\right.$

$⇒ \left[\begin{matrix} x=8 - \frac{1}{2}\\ x+\frac{1}{2} = 9\\ x=14 \end{matrix}\right.$

$⇒ \left[\begin{matrix} x=\frac{16}{2} - \frac{1}{2}\\ x= \frac{18}{2} - \frac{1}{2} \\ x=14 \end{matrix}\right.$

$⇒ \left[\begin{matrix} x=\frac{15}{2}\\ x= \frac{17}{2} \\ x=14 \end{matrix}\right.$

Vậy $ x ∈ { \frac{15}{2} ; \frac{17}{2}; 14 } $

$ c) ( x^{2} + 2x )^{2} - 2x^{2} - 4x = 3 $

$ ⇔ ( x^{2} + 2x )^{2} - 2x ( x + 2 ) = 4 - 1 $

$ ⇔ ( x^{2} + 2x )^{2} - 2x ( x + 2 ) + 1 = 4  $

$ ⇔ ( x^{2} + 2x - 1 )^{2} = 4  $

$ ⇔ x^{2} + 2x - 1  = \pm2  $

$⇒ \left[\begin{matrix} x^{2} + 2x - 1  =2\\ x^{2} + 2x - 1  =-2 \end{matrix}\right.$

$⇒ \left[\begin{matrix} x^{2} + 2x - 1  =2\\ x^{2} + 2x - 1  =-2 \end{matrix}\right.$

$⇒ \left[\begin{matrix} x^{2} - x + 3x  - 3 = 0\\ x^{2} + 2 . x + 1^{2} = 0 \end{matrix}\right.$

$⇒ \left[\begin{matrix} x ( x - 1 ) + 3 ( x - 1 ) = 0\\  ( x + 1)^{2} = 0 \end{matrix}\right.$

$⇒ \left[\begin{matrix} ( x - 1 ) ( x + 3 )= 0\\  ( x + 1)^{2} = 0 \end{matrix}\right.$

$⇒ \left[\begin{matrix}x - 1 = 0\\ x + 3 = 0 \\ ( x + 1 )^{2} = 0\end{matrix}\right.$

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

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

Vậy $ x ∈ { 1 ; - 3; - 1 } $