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

Toán Lớp 8: B1 : Tìm x biết 5(x+3) – 2x(x+3) = 0 B2 : Rút gọn A = $\frac{8-x}{(x+2)(x-3)}$+$\frac{2}{x+2}$ Giúp mình với, mình cảm ơn nhiều <3

Toán Lớp 8: B1 : Tìm x biết
5(x+3) – 2x(x+3) = 0
B2 : Rút gọn
A = $\frac{8-x}{(x+2)(x-3)}$+$\frac{2}{x+2}$
Giúp mình với, mình cảm ơn nhiều <3

Comments ( 2 )

  1. B1:
    5(x + 3) – 2x(x + 3) = 0
    (x + 3)(5 – 2x) = 0
    TH1: x + 3 = 0
            x = 0 – 3 = -3
    TH2: 5 – 2x = 0
                  2x = 5 – 0 = 5
                    x = 5 : 2 = $\frac{5}{2}$ 
    KL: x ∈ {-3; $\frac{5}{2}$}
    B2:
    A = $\frac{8-x}{(x+2)(x-3)}$ + $\frac{2}{x+2}$
    = $\frac{8-x}{(x+2)(x-3)}$ + $\frac{2(x-3)}{(x+2)(x-3)}$
    = $\frac{8-x}{(x+2)(x-3)}$ + $\frac{2x-6}{(x+2)(x-3)}$
    = $\frac{8-x+2x-6}{(x+2)(x-3)}$
    = $\frac{x+2}{(x+2)(x-3)}$
    = $\frac{1}{x-3}$

  2. Giải đáp:
     $B1:$
    $5.(x+3)-2x.(x+3)=0$
    $⇔ (5-2x).(x+3)=0$
    $⇔$ \(\left[ \begin{array}{l}5-2x=0\\x+3=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}2x=5\\x=-3\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=-3\end{array} \right.\) 
    $Vậy$ $\text{S={5/2;-3}}$
    $B2:$
    $A= \dfrac{8-x}{(x+2).(x-3)}+\dfrac{2}{x+2}$
    $= \dfrac{8-x}{(x+2).(x-3)}+\dfrac{2.(x-3)}{(x+2).(x-3)}$
    $= \dfrac{8-x+2x-6}{(x+2).(x-3)}$
    $= \dfrac{x+2}{(x+2).(x-3)}$
    $= \dfrac{1}{x-3}$
     

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