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

Toán Lớp 8: tìm x ạ ==- (đăng lại mong đc trl hay hơn ạ) 9x^2-4-2(3x-2)=0 3x^2-7x+4=0

Tháng Ba 15, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: tìm x ạ ………………………. (đăng lại mong đc trl hay hơn ạ)
9x^2-4-2(3x-2)=0
3x^2-7x+4=0

Comments ( 2 )

  1. thanoanghha
    thanoanghha
    15 Tháng Ba, 2022 at 9:45 sáng
    Reply
    9x^2-4-2(3x-2)=0
    => 9x^2 – 4 – 6x + 4 = 0
    => 9x^2 – 6x = 0
    => x(9x-6)=0
    => $\left[\begin{matrix} x=0\\ 9x-6=0\end{matrix}\right.$
    => $\left[\begin{matrix} x=0\\ x=2/3\end{matrix}\right.$
    Vậy x in {0;2/3}
    3x^2-7x+4=0
    => (3x^2-4x)+(-3x+4) = 0
    => x(3x-4)-(3x-4) = 0
    => (3x-4)(x-1) = 0
    => $\left[\begin{matrix} 3x-4=0\\ x-1=0\end{matrix}\right.$
    => $\left[\begin{matrix} x= \dfrac{4}{3}\\ x=1\end{matrix}\right.$
    Vậy x in {4/3;1}

  2. minhhaanh
    minhhaanh
    15 Tháng Ba, 2022 at 9:45 sáng
    Reply
    1/
    9x² – 4 – 2(3x – 2) = 0
    ⇔ (3x – 2)(3x + 2) – 2(3x – 2) = 0
    ⇔ (3x – 2)(3x + 2 – 2) = 0
    ⇔ (3x – 2)3x = 0
    ⇔ \(\left[ \begin{array}{l}3x-2=0\\3x=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=\frac{2}{3}\\x=0\end{array} \right.\)
    Vậy x = 0 hoặc x = $\frac{2}{3}$
    2/
    3x²  -7x + 4 = 0
    ⇔ 3x² – 3x – 4x + 4 = 0
    ⇔ 3x(x – 1) – 4(x – 1) = 0
    ⇔ (3x – 4)(x – 1) = 0
    ⇔ \(\left[ \begin{array}{l}3x-4=0\\x-1=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=\frac{4}{3}\\x=1\end{array} \right.\)
    Vậy x = 1 hoặc x = $\frac{4}{3}$

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

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