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

Toán Lớp 9: B1: giả bpt sau a) (x-1/2)(2x+5)=0 b) 15-7x = 9-3x c) 3x-1/x-1 – 2x+5/x-3 = 1

Toán Lớp 9: B1: giả bpt sau
a) (x-1/2)(2x+5)=0
b) 15-7x = 9-3x
c) 3x-1/x-1 – 2x+5/x-3 = 1

Comments ( 2 )

  1. a)
    (x – 1/2)(2x + 5) = 0
    ⇒ \(\left[ \begin{array}{l}x – 1/2 = 0\\2x + 5 = 0\end{array} \right.\) 
    ⇒ \(\left[ \begin{array}{l}x = 1/2\\x = -5/2\end{array} \right.\) 
    Vậy  S = {1/2 ; -5/2}
    b)
    15 – 7x = 9 – 3x
    ⇒ 15 – 9 = -3x + 7x
    ⇒ 6 = 4x
    ⇒ x = 3/2
    Vậy  S = {3/2}
    c)
    (3x – 1)/(x – 1) – (2x + 5)/(x – 3) = 1
    ⇔ ((3x – 1)(x – 3))/((x – 1)(x – 3)) – ((2x + 5)(x – 1))/((x – 1)(x – 3)) – 1 = 0
    ⇔ (3x^2 – 10x + 3 – 2x^2 – 3x + 5)/((x – 1)(x – 3)) – ((x – 1)(x – 3))/((x – 1)(x – 3)) = 0
    ⇔ (3x^2 – 10x + 3 – 2x^2 – 3x + 5 – (x – 1)(x – 3))/((x – 1)(x – 3)) = 0
    ⇔ x^2 – 13x + 8 – x^2 + 4x – 3 = 0
    ⇔ -9x + 5 = 0
    ⇔ -9x/-5
    ⇒ x = 5/9
    Vậy  S = {5/9}

  2. Bài $1$.
    $a$) (x-1/2)(2x+5) = 0
    ⇒ \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{5}{2}\end{array} \right.\) 
       Vậy $x$ $∈$ {1/2; -5/2}.
    $b$) $15-7x = 9-3x$
    $⇔ -7x + 3x = 9 – 15$
    $⇔ -4x = -6$
    $⇔ x = \dfrac{3}{2}$
      Vậy x=3/2.
    $c$) {3x-1}/{x-1} – {2x+5}/{x-3}  = 1  ($x \neq 1;3$)
    ⇔ {(3x-1)(x-3) – (x-1)(2x+5)}/{(x-1)(x-3)} – 1 = 0
    ⇔ {3x^2 – 10x+3 – 2x^2 – 3x + 5 – (x-1)(x-3)}/{(x-1)(x-3)} = 0
    ⇔ x^2 – 13x +8 – x^2 + 4x – 3 = 0
    ⇔ -9x +5 = 0
    ⇔ x = 5/9
       Vậy x=5/9.

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