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

Toán Lớp 8: Tìm x x^3– 9x = 0 b) (x + 1)^2 – (7x + 1)^2 = 0

Tháng Bảy 7, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: Tìm x
x^3– 9x = 0
b) (x + 1)^2 – (7x + 1)^2 = 0

Comments ( 2 )

  1. dananhthumai
    dananhthumai
    7 Tháng Bảy, 2022 at 7:05 sáng
    Reply
    $\text{a) $x^{3}$-9x=0}$
    $\text{$\Leftrightarrow$ x($x^{2}$-9)=0}$
    $\text{$\Leftrightarrow$ x(x-3)(x+3)=0}$
    $\Leftrightarrow$ $\left[\begin{matrix} x=0\\ x-3=0\\ x+3=0\end{matrix}\right.$
    $\Leftrightarrow$ $\left[\begin{matrix} x=0\\ x=3\\ x=-3\end{matrix}\right.$
    $\text{b) $(x+1)^{2}$-$(7x+1)^{2}$=0}$
    $\text{$\Leftrightarrow$ (x+1-7x-1)(x+1+7x+1)=0}$
    $\text{$\Leftrightarrow$ (-6x)(8x+2)=0}$
    $\Leftrightarrow$ $\left[\begin{matrix} -6x=0\\ 8x+2=0\end{matrix}\right.$
    $\Leftrightarrow$ $\left[\begin{matrix} x=0\\ 8x=-2\end{matrix}\right.$
    $\Leftrightarrow$ $\left[\begin{matrix} x=0\\ x=\dfrac{-1}{4}\end{matrix}\right.$

  2. thaolanphuobg
    thaolanphuobg
    7 Tháng Bảy, 2022 at 7:05 sáng
    Reply
    a) x^3 – 9x = 0
    ⇔ x (x^2-9) = 0
    ⇔ x (x-3) (x+3) = 0
    +) x = 0
    +) x – 3 = 0
    ⇔ x = 3
    +) x + 3 = 0
    ⇔ x = -3
    Vậy . . .
    b) (x+1)^2 –  (7x+1)^2 = 0
    ⇔ (x^2+2x+1) – (49x^2+14x+1) = 0
    ⇔ x^2 + 2x + 1 – 49x^2 – 14x – 1 = 0
    ⇔ -48x^2 – 12x = 0
    ⇔ – 12x (4x+1) = 0
    +) – 12x = 0
    ⇔ x = 0
    +) 4x + 1 = 0
    ⇔ 4x = -1
    ⇔ x = -1/4
    Vậy . . .
     

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

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