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

Toán Lớp 8: a)(x^2-9)-(x+2)(x+3)=0 b)(2x-7)^2=(2x-7) c)3x^3-7x=0 tìm x ạ đúng vote 5 sao

Home/ Toán Lớp 8: a)(x^2-9)-(x+2)(x+3)=0 b)(2x-7)^2=(2x-7) c)3x^3-7x=0 tìm x ạ đúng vote 5 sao

Toán Lớp 8: a)(x^2-9)-(x+2)(x+3)=0 b)(2x-7)^2=(2x-7) c)3x^3-7x=0 tìm x ạ đúng vote 5 sao

Tháng Mười Một 13, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: a)(x^2-9)-(x+2)(x+3)=0
b)(2x-7)^2=(2x-7)
c)3x^3-7x=0
tìm x ạ đúng vote 5 sao

Comments ( 2 )

  1. lanthuhoa
    lanthuhoa
    13 Tháng Mười Một, 2022 at 8:16 sáng
    Reply
    a,(x^2-9)-(x+2)(x+3)=0
    ⇔(x-3)(x+3)-(x+2)(x+3)=0
    ⇔(x+3)(x-3-x-2)=0
    ⇔(x+3).(-5)=0
    ⇔x+3=0
    ⇔x=-3
     Vậy: S={-3}
    b,(2x-7)^2=2x-7
    ⇔(2x-7)^2-(2x-7)=0
    ⇔(2x-7)(2x-7-1)=0
    ⇔(2x-7)(2x-8)=0
    ⇔\(\left[ \begin{array}{l}2x-7=0\\2x-8=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}2x=7\\2x=8\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=4\end{array} \right.\) 
    Vậy: S={\frac{7}{2};4}
    c,3x^3-7x=0
    ⇔x(3x^2-7)=0
    ⇔\(\left[ \begin{array}{l}x=0\\3x^2-7=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=0\\3x^2=7\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=0\\x^2=\dfrac{7}{3}\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=0\\x=\sqrt{\dfrac{7}{3}}\\x=-\sqrt{\dfrac{7}{3}}\end{array} \right.\) 
    Vậy: S={0;\sqrt{\frac{7}{3}};-\sqrt{\frac{7}{3}}}

  2. thanoanghha
    thanoanghha
    13 Tháng Mười Một, 2022 at 8:16 sáng
    Reply
    Giải đáp:
     a, (x^2 – 9) – (x + 2)(x + 3) = 0
    ⇔ (x^2 – 3^2) – (x + 2)(x + 3) = 0
    ⇔ (x – 3)(x + 3) – (x + 2)(x + 3) = 0
    ⇔ (x + 3)(x – 3 – x – 2) = 0
    ⇔ (x + 3). (-5) = 0
    ⇔ x + 3 = 0
    ⇒ x =0 – 3
    ⇒ x = -3
    Vậy x = -3
    b, (2x – 7)^2 = (2x – 7)
    ⇔ (2x – 7)^2 – (2x – 7) = 0
    ⇔ (2x – 7)(2x – 7 – 1) = 0
    ⇔ (2x – 7)(2x – 8) = 0
    ⇔ (2x – 7)(x – 4).2 = 0
    ⇒ $\left[\begin{matrix} 2x – 7 = 0\\ x – 4 = 0\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} 2x = 7\\ x = 4\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x = \frac{7}{2}\\ x = 4\end{matrix}\right.$
    Vậy x = 7/2 hoặc x = 4
    c, 3x^3 – 7x = 0
    ⇔ x(3x^2 – 7) = 0
    ⇒ $\left[\begin{matrix} x = 0\\ 3x^2 – 7 = 0\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x = 0\\ 3x^2 = 7\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x = 0\\ x^2 = \dfrac{7}{3}\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x = 0\\ x^2 = (\sqrt{\dfrac{7}{3}})^2\\x^2 = (-\sqrt{\dfrac{7}{3}})^2\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x = 0\\ x = \sqrt{\dfrac{7}{3}}\\x^2 = -\sqrt{\dfrac{7}{3}}\end{matrix}\right.$
    Vậy x = 0, x =$\sqrt{\dfrac{7}{3}}$ hoặc x = $-\sqrt{\dfrac{7}{3}}$

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

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