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

Toán Lớp 8: tìm x 5x + x^2 = 0 2(x+5) – x^2 – 5x = 0

Home/ Toán Lớp 8: tìm x 5x + x^2 = 0 2(x+5) – x^2 – 5x = 0

Toán Lớp 8: tìm x 5x + x^2 = 0 2(x+5) – x^2 – 5x = 0

Tháng Ba 17, 2023 Toán học 2 Comments 0 views

Toán Lớp 8: tìm x
5x + x^2 = 0
2(x+5) – x^2 – 5x = 0

Comments ( 2 )

  1. hoaiphuong85
    hoaiphuong85
    17 Tháng Ba, 2023 at 11:33 sáng
    Reply
    Giải đáp + giải thích các bước giải:
    5x + x^2 = 0
    <=> x(x + 5) = 0
    <=> \(\left[ \begin{array}{l}x=0\\x+5=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\) 
    Vậy x \in {0 ; -5}
    2(x + 5) – x^2 – 5x = 0
    <=> 2(x + 5) – (x^2 + 5x) = 0
    <=> 2(x + 5) – x(x + 5) = 0
    <=> \(\left[ \begin{array}{l}2 – x = 0\\x+5=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x= 2\\x=-5\end{array} \right.\) 
    Vậy x \in {2 ; -5}
     

  2. behocgioitoan
    behocgioitoan
    17 Tháng Ba, 2023 at 11:33 sáng
    Reply
    1)
    $5x+x^2=0$
    $⇔ x^2+5x=0$
    $⇔x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4.1.0}}{2.1}$ 
    $⇔x_{1,\:2}=\frac{-5\pm \:5}{2.1}$
    $⇔x_1=\frac{-5+5}{2.1},\:x_2=\frac{-5-5}{2.1}$
    \(⇔\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\) 
    2) 
    $2\left(x+5\right)-x^2-5x=0$
    $⇔ -x^2-3x+10=0$
    $⇔x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\left(-1\right)\cdot \:10}}{2\left(-1\right)}$
    $⇔x_{1,\:2}=\frac{-\left(-3\right)\pm \:7}{2\left(-1\right)}$
    $⇔x_1=\frac{-\left(-3\right)+7}{2\left(-1\right)},\:x_2=\frac{-\left(-3\right)-7}{2\left(-1\right)}$
    \(⇔\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\) 

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

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