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222-9+11+12:2*14+14 = ? ( )

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

Toán Lớp 8: 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

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    $ 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 } $

  2. Giải đáp:+ Lời giải và giải thích chi tiết: 
    a) x.( 2x – 3 )- 2. ( 3- 2x )=0
    ⇔ x.( 2x – 3 )+ 2.( 2x – 3 )=0
    ⇔ ( 2x – 3 )( x + 2 )= 0
    ⇒ 2x – 3 = 0 hoặc x + 2= 0
    th1:  2x – 3= 0            Th2:   x + 2 = 0
              2x    = 0+3                    x    = 0 – 2
               2x    = 3                          x  = -2
                 x= 3/2
    Vậy x ∈ { 3/2 ; -2 }
     b) ( x+1/2)² – ( x + 1/2). ( x+ 6) =8
    ⇔ ( x + 1/2 ).( x+ 1/2 – x – 6 ) = 8
    ⇔ ( x + 1/2 ). (-11/2)  =8
    ⇔ x + 1/2  = 8 :  (-11/2)
    ⇔ x +1/2 = -16/11
    ⇔ x    = -16/11 – 1/2
    ⇔ x    = -43/22
    vậy x= -43/22.
    c)

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222-9+11+12:2*14+14 = ? ( )