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

Toán Lớp 8: (3x-1)^2-(2x+3)^2=0 2020x^2-2021x+1=0

Tháng Mười 3, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: (3x-1)^2-(2x+3)^2=0
2020x^2-2021x+1=0

Comments ( 2 )

  1. haianhtu97
    haianhtu97
    3 Tháng Mười, 2022 at 7:48 chiều
    Reply
    .
    .    $(3x-1)^2-(2x+3)^2=0$
    ⇔ $\left[(3x-1)+(2x+3)\right].\left[(3x-1)-(2x+3)\right]=0$
    ⇔ $\left(5x+2\right).\left(x-4\right)=0$
    ⇔ $\left[\begin{array}{l}5x+2=0\\x-4=0\end{array}\right.$
    ⇔ $\left[\begin{array}{l}x=-\frac{2}{5}\\x=4\end{array}\right.$
    $\text{- Vậy $S=\left\{-\frac{2}{5};0\right\}$}$
    $\text{}$
        $2020x^2-2021x+1=0$
    ⇔ $2020x^2-2020x-x+1=0$
    ⇔ $2020x(x-1)-(x-1)=0$
    ⇔ $(x-1)(2020x-1)=0$
    ⇔ $\left[\begin{array}{l}x-1=0\\2020x-1=0\end{array}\right.$
    ⇔ $\left[\begin{array}{l}x=1\\x=\frac{1}{2020}\end{array}\right.$
    $\text{- Vậy $S=\left\{\frac{1}{2020};1\right\}$}$

  2. lanhuongthu
    lanhuongthu
    3 Tháng Mười, 2022 at 7:48 chiều
    Reply
    Giải đáp + Lời giải và giải thích chi tiết :
    ( 3x – 1 )^2 – ( 2x + 3 )^2 = 0
    ⇔ [ 3x – 1 – ( 2x + 3 ) ] . [ 3x – 1 + ( 2x + 3 ) ] = 0
    ⇔ ( 3x – 1 – 2x – 3 ) . ( 3x – 1 + 2x + 3 ) = 0
    ⇔ ( x – 4 ) . ( 5x + 2 ) = 0
    ⇔ \(\left[ \begin{array}{l}x-4=0\\5x+2=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=4\\x=-\frac{2}{5}\end{array} \right.\)  
    Vậy x = 4 ; x = – 2/5
    .
    2020x^2 – 2021x + 1 = 0
    ⇔ 2020x^2 – x – 2020x + 1 = 0
    ⇔ x ( 2020x – 1 ) – ( 2020x – 1 ) = 0
    ⇔ ( x – 1 ) . ( 2020x – 1 ) = 0
    ⇔ \(\left[ \begin{array}{l}x-1=0\\2020x-1=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=1\\x=\frac{1}{2020}\end{array} \right.\) 
    Vậy x = 1 ; x = 1/2020
    $# Mon $
    $# Xin Hay Nhất $

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

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