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

Toán Lớp 9: Tìm x biết 4.(x+1)+3.(x+3)=7 x^2-3x=0 2020x.(x-2019)+x_2019=0

Tháng Hai 18, 2022 Toán học 2 Comments 0 views

Toán Lớp 9: Tìm x biết
4.(x+1)+3.(x+3)=7
x^2-3x=0
2020x.(x-2019)+x_2019=0

Comments ( 2 )

  1. haianhtu97
    haianhtu97
    18 Tháng Hai, 2022 at 8:52 sáng
    Reply
    Answer
    4 . (x + 1) + 3 . (x + 3) = 7
    => 4 . x + 4 . 1 + 3 . x + 3 . 3 = 7
    => 4x + 4 + 3x + 9 = 7
    => (4x + 3x) + (4 + 9) = 7
    => 7x + 13 = 7
    => 7x = 7 – 13
    => 7x = -6
    => x = -6 : 7
    => x = -6/7
    Vậy S = {-6/7}
    _______________________
    x^2 – 3x = 0
    => x . (x – 3) = 0
    => $\left[\begin{matrix} x = 0\\ x – 3 = 0\end{matrix}\right.$
    => $\left[\begin{matrix} x = 0\\ x = 0 + 3\end{matrix}\right.$
    => $\left[\begin{matrix} x = 0\\ x = 3\end{matrix}\right.$
    Vậy S = {0 ; 3}
    __________________________
    2020x . (x – 2019) + x – 2019 = 0
    => 2020x . (x – 2019) + (x – 2019) = 0
    => (2020x + 1) . (x – 2019) = 0
    => $\left[\begin{matrix} 2020x + 1 = 0\\ x – 2019 = 0\end{matrix}\right.$
    => $\left[\begin{matrix} 2020x = 0 – 1\\ x = 0 + 2019\end{matrix}\right.$
    => $\left[\begin{matrix} 2020x = – 1\\ x = 2019\end{matrix}\right.$
    => $\left[\begin{matrix} x = – 1 : 2020\\ x = 2019\end{matrix}\right.$
    => $\left[\begin{matrix} x = \dfrac{-1}{2020}\\ x = 2019\end{matrix}\right.$
    Vậy S = {-1/2020 ; 2019}

  2. ngochuong2k2
    ngochuong2k2
    18 Tháng Hai, 2022 at 8:51 sáng
    Reply
    Giải đáp+Lời giải và giải thích chi tiết:
    4(x+1)+3(x+3)=7
    <=> 4x+4+3x+9=7
    <=> 7x+13=7
    <=> 7x=-6
    <=> x=-6/7
    Vậy x=-6/7
    x^2-3x=0
    <=> x(x-3)=0
    <=> [(x=0),(x-3=0):}
    <=> [(x=0),(x=3):}
    Vậy x \in {0;3}
    2020x(x-2019)+x-2019=0
    <=> 2020x(x-2019)+(x-2019)=0
    <=> (2020x+1)(x-2019)=0
    <=> [(2020x+1=0),(x-2019=0):}
    <=> [(x=-1/2020),(x=2019):}
    Vậy x \in {-1/2020;2019} 

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

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