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

Toán Lớp 8: tìm x, biết: a) 16x ² – 25 = 0 b)x ² 2021x + 2020 = 0

Home/ Toán Lớp 8: tìm x, biết: a) 16x ² – 25 = 0 b)x ² 2021x + 2020 = 0

Toán Lớp 8: tìm x, biết: a) 16x ² – 25 = 0 b)x ² 2021x + 2020 = 0

Tháng Tư 30, 2023 Toán học 2 Comments 0 views

Toán Lớp 8: tìm x, biết:
a) 16x ² – 25 = 0
b)x ² 2021x + 2020 = 0

Comments ( 2 )

  1. quynhthanhnghi
    quynhthanhnghi
    30 Tháng Tư, 2023 at 12:59 sáng
    Reply
    a) 16x² – 25 = 0
    ⇔ ( 4x )² – 5² = 0
    ⇔ ( 4x – 5 ) ( 4x + 5 ) = 0
    ⇔ $\left[\begin{matrix} 4x – 5 = 0 \\ 4x + 5 = 0 \end{matrix}\right.$
    ⇔ $\left[\begin{matrix} x = \dfrac{5}{4}\\ x = \dfrac{-5}{4}\end{matrix}\right.$
      Vậy x = $\dfrac{5}{4}$ hoặc x = $\dfrac{-5}{4}$
    b)  x² – 2021x + 2020 = 0
    ⇔ x² – x – 2020x + 2020 = 0
    ⇔ ( x² – x ) – ( 2020x – 2020 ) = 0
    ⇔ x ( x – 1 ) – 2020 ( x – 1 ) = 0
    ⇔ ( x – 1 ) ( x – 2020 ) = 0
    ⇔ $\left[\begin{matrix} x – 1 = 0 \\ x – 2020 = 0 \end{matrix}\right.$
    ⇔ $\left[\begin{matrix} x = 1 \\ x = 2020\end{matrix}\right.$
     Vậy x = 1 hoặc x = 2020

  2. thuminhthong
    thuminhthong
    30 Tháng Tư, 2023 at 12:59 sáng
    Reply
    Gửi bạn:
    $a,$ $16x^2-25=0$
    $⇒(4x)^2-(5)^2=0$
    $⇒(4x-5)(4x+5)=0$
    $⇒$ $\left[\begin{matrix} 4x-5=0\\ 4x+5=0\end{matrix}\right.$
    $⇒$ $\left[\begin{matrix} x= \dfrac{5}{4} \ x=-\dfrac{5}{4} \end{matrix}\right.$
    Vậy $x=±\dfrac{5}{4}$
    $b,$ $x^2-2021x+2020=0$
    $x^2-x-2020x+2020=0$
    $⇒x(x-1)-2020(x-1)=0$
    $⇒(x-1)(x-2020)=0$
    $⇒$ $\left[\begin{matrix} x-1=0\\ x-2020=0\end{matrix}\right.$
    $⇒$ $\left[\begin{matrix} x=1\\ x=2020\end{matrix}\right.$
    Vậy $x=1$ hoặc $x=2020$

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

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