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

Toán Lớp 8: Tìm x: a)x2+10x=0 b)2x(x-1)-3(1-x)=0 c)4×2-9=0

Tháng Mười Một 26, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: Tìm x: a)x2+10x=0
b)2x(x-1)-3(1-x)=0
c)4×2-9=0

Comments ( 2 )

  1. tuyetnhungthu
    tuyetnhungthu
    26 Tháng Mười Một, 2022 at 1:17 sáng
    Reply
    a, x^2 + 10x = 0
        x(x + 10) = 0
    => \(\left[ \begin{array}{l}x=0\\x+10=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=0\\x=-10\end{array} \right.\) 
       Vậy x ∈ {-10; 0}
    b, 2x(x – 1) – 3(1 – x) = 0
        2x(x – 1) + 3(x – 1) = 0
        (2x + 3)(x – 1)        = 0
    => \(\left[ \begin{array}{l}2x+3=0\\x-1=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=\frac{-3}{2}\\x=1\end{array} \right.\) 
        Vậy x ∈ {-3/2; 1}
    c, 4x^2 – 9 = 0
       (2x)^2 – 3^2 = 0
       (2x – 3)(2x + 3) = 0
    => \(\left[ \begin{array}{l}2x-3=0\\2x+3=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=\frac{3}{2}\\x=-\frac{3}{2}\end{array} \right.\) 
        Vậy x ∈ {±3/2}

  2. phuongthaongoc
    phuongthaongoc
    26 Tháng Mười Một, 2022 at 1:16 sáng
    Reply
    $\text{a) }$$x^{2}+10x=0$
    $\text{x.(x+10)=0}$
    \(\left[ \begin{array}{l}x=0\\x+10=0\end{array} \right.\)
    \(\left[ \begin{array}{l}x=0\\x=-10\end{array} \right.\)
    ___________________________________
    $\text{b) 2x.(x-1)-3.(1-x)=0}$
    $\text{2x.(x-1)+3.(x-1)=0}$
    $\text{(x-1).(2x+3)=0}$
    \(\left[ \begin{array}{l}x-1=0\\2x+3=0\end{array} \right.\)
    \(\left[ \begin{array}{l}x=1\\x=\dfrac{-3}{2}\end{array} \right.\)
    ___________________________________
    \text{c) 4}$x^{2}-9=0$
    \text{(2x)}$^{2}-$ $3^{2}=0$
    \text{(2x-3).(2x+3)=0}
    \(\left[ \begin{array}{l}2x-3=0\\2x+3=0\end{array} \right.\)
    \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{-3}{2}\end{array} \right.\)
    $\text{-Áp dụng HĐT hiệu 2 bình phương:}$$A^{2}-$ $B^{2}=(A-B).(A+B)$
    $\text{#mct}$

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

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