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

Toán Lớp 8: Tìm x: A)(x-2)^2-1=0 B)x^2-9-8/9*x^2=0 C)(3x-2)^2-(2x+3)^2=5(x-4)*(x+4)

Toán Lớp 8: Tìm x:
A)(x-2)^2-1=0
B)x^2-9-8/9*x^2=0
C)(3x-2)^2-(2x+3)^2=5(x-4)*(x+4)

Comments ( 2 )

  1. $Quangphuc347$????
    ta có :
    a,  $(x-2)^2$ $-$ $1$ $=$ $0$
    ⇒ $(x-2)^2$ $=$ $1$
    ⇒ $\left[\begin{matrix} x-2=1\\ x+2=-1\end{matrix}\right.$
    ⇔ $\left[\begin{matrix} x = 3\\ x = -3\end{matrix}\right.$
    b, $x^2$ $-$ $9$ $-$$\frac{8}{9}$ x $x^2$ $=$ $0$
    ⇒ $\frac{x^2}{9}$ – $9$ $=$ $0$
    ⇒ $\frac{x^2}{9}$  $=$ $-9$
    ⇒  $x^2$ $=$ $81$
    ⇒$\left[\begin{matrix} x = 9\\ x = -9\end{matrix}\right.$
    c, $3x-2^2$ $-$ ($2x+3^2$) $=$ $5(x-4)(x+4)$
    đợi tui suy nghĩ câu này nha $:)$.

  2. a) (x-2)^2-1=0
    ⇔(x-2-1)(x-2+1)=0
    ⇔(x-3)(x-1)=0
    ⇔\(\left[ \begin{array}{l}x-3=0\\x-1=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\) 
    Vậy S={3;1}
    b) x^2-9-\frac{8x^2}{9}=0
    ⇔\frac{x^2}{9}-9=0
    ⇔\frac{x^2}{9}=9
    ⇔x^2=81
    ⇔\(\left[ \begin{array}{l}x=9\\x=-9\end{array} \right.\) 
    Vậy S={9;-9}
    c) (3x-2)^2-(2x+3)^2=5(x-4)(x+4)
    ⇔9x^2-12x+4-4x^2-12x-9-5x^2+80=0
    ⇔-24x+75=0
    ⇔-24x=-75
    ⇔x=\frac{75}{24}=\frac{25}{8}
    Vậy S={\frac{25}{8}}

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