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

Toán Lớp 8: a) x( x-2 ) + x – 2 = 0 b) 5x( x-3 ) – x+3 = 0 c) (3x + 5)(4 – 3x) = 0 d) 3x(x – 7) – 2(x – 7) = 0 e) 7x^2 – 28 = 0 f) (2x + 1) + x

Home/ Toán Lớp 8: a) x( x-2 ) + x – 2 = 0 b) 5x( x-3 ) – x+3 = 0 c) (3x + 5)(4 – 3x) = 0 d) 3x(x – 7) – 2(x – 7) = 0 e) 7x^2 – 28 = 0 f) (2x + 1) + x

Toán Lớp 8: a) x( x-2 ) + x – 2 = 0 b) 5x( x-3 ) – x+3 = 0 c) (3x + 5)(4 – 3x) = 0 d) 3x(x – 7) – 2(x – 7) = 0 e) 7x^2 – 28 = 0 f) (2x + 1) + x

Tháng Mười Hai 23, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: a) x( x-2 ) + x – 2 = 0
b) 5x( x-3 ) – x+3 = 0
c) (3x + 5)(4 – 3x) = 0
d) 3x(x – 7) – 2(x – 7) = 0
e) 7x^2 – 28 = 0
f) (2x + 1) + x(2x + 1) = 0
g) 2x^3 – 50x = 0
h)2x(3x-5)-(5-3x)=0
giúp ạ

Comments ( 2 )

  1. thanhthuythu
    thanhthuythu
    23 Tháng Mười Hai, 2022 at 11:37 chiều
    Reply
    a)x(x-2)+x-2=0
    ⇔x(x-2)+(x-2)=0
    ⇔(x-2)(x+1)=0
    ⇔$\left[\begin{matrix} x-2=0\\ x+1=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=2\\ x=-1\end{matrix}\right.$
    Vậy S={2;-1}
    b)5x(x-3)-x+3=0
    ⇔5x(x-3)-(x-3)=0
    ⇔(x-3)(5x-1)=0
    ⇔$\left[\begin{matrix} x-3=0\\ 5x-1=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=3\\ x=\dfrac{1}{5}\end{matrix}\right.$
    Vậy S={3;1/5}
    c)(3x+5)(4-3x)=0
    ⇔$\left[\begin{matrix} 3x+5=0\\ 4-3x=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} 3x=-5\\ 3x=4\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=-\dfrac{5}{3}\\ x=\dfrac{4}{3}\end{matrix}\right.$
     Vậy S={-5/3;4/3}
    d)3x(x-7)-2(x-7)=0
    ⇔(x-7)(3x-2)=0
    ⇔$\left[\begin{matrix} x-7=0\\ 3x-2=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=7\\ x=\dfrac{2}{3}\end{matrix}\right.$
    Vậy S={7;2/3}
    e)7x²-28=0
    ⇔7(x²-4)=0
    ⇔7(x²-2²)=0
    ⇔7(x+2)(x-2)=0
    ⇔(x+2)(x-2)=0
    ⇔$\left[\begin{matrix} x+2=0\\ x-2=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=-2\\ x=2\end{matrix}\right.$
    Vậy S={-2;2}
    f)(2x+1)+x(2x+1)=0
    ⇔(2x+1)(x+1)=0
    ⇔$\left[\begin{matrix} 2x+1=0\\ x+1=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=-\dfrac{1}{2}\\ x=-1\end{matrix}\right.$
    Vậy S={-1/2;-1}
    g)2x³-50x=0
    ⇔2x(x²-25)=0
    ⇔2x(x²-5²)=0
    ⇔2x(x+5)(x-5)=0
    ⇔$\left[\begin{matrix} 2x=0\\x+5=0\\ x-5=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=0\\x=-5\\ x=5\end{matrix}\right.$
    Vậy S={0;-5;5}
    h)2x(3x-5)-(5-3x)=0
    ⇔2x(3x-5)+(3x-5)=0
    ⇔(3x-5)(2x+1)=0
    ⇔$\left[\begin{matrix} 3x-5=0\\ 2x+1=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} 3x=5\\ 2x=-1\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=\dfrac{5}{3}\\ x=-\dfrac{1}{2}\end{matrix}\right.$
    Vậy S={5/3;-1/2}

  2. behocgioitoan
    behocgioitoan
    23 Tháng Mười Hai, 2022 at 11:37 chiều
    Reply
    a)x( x-2 ) + x – 2 = 0
    <=>x(x-2)+(x-2)=0
    <=>(x+1)(x-2)=0
    <=>\(\left[ \begin{array}{l}x+1=0\\x-2=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\)
    Vậy S={-1;2}
    b) 5x( x-3 ) – x+3 = 0
    <=>5x( x-3 ) – (x-3)=0
    <=>(x-3)(5x-1)=0
    <=>\(\left[ \begin{array}{l}x-3=0\\5x-1=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=3\\5x=1\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=3\\x=1/5\end{array} \right.\)
    Vậy S={3;1/5}
    c)(3x + 5)(4 – 3x) = 0
    <=>\(\left[ \begin{array}{l}3x+5=0\\4-3x=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}3x=-5\\-3x=-4\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=-5/3\\x=3/4\end{array} \right.\)
    Vậy S={-5/3;3/4}
    d)3x(x – 7) -2(x – 7) = 0
    <=>(x-7)(3x-2)=0
    <=>\(\left[ \begin{array}{l}x-7=0\\3x-2=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=7\\3x=2\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=7\\x=2/3\end{array} \right.\)
    Vậy S={7;2/3}
    e)7x^2 – 28 = 0
    <=>7(x^2-4)=0
    <=>7(x^2-2^2)=0
    <=>7(x-2)(x+2)=0
    <=>(x-2)(x+2)=0
    <=>\(\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
    Vậy S={-2;+2}
    f)(2x + 1) + x(2x + 1) = 0
    <=>(2x+1)(x+1)=0
    <=>\(\left[ \begin{array}{l}2x+1=0\\x+1=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}2x=-1\\x=-1\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=-1/2\\x=-1\end{array} \right.\)
    Vậy S={-1/2;-1}
    g)2x^3 – 50x = 0
    <=>2x(x^2-25)=0
    <=>2x(x-5)(x+5)=0
    <=>2x=0 hoặc x-5=0 hoặc x+5=0
    <=>x=0 hoặc x=5 hoặc x=-5
    Vậy S={0;5;-5}
    h)2x(3x-5)-(5-3x)=0
    <=>2x(3x-5)+(3x-5)=0
    <=>(3x-5)(2x+1)=0
    <=>\(\left[ \begin{array}{l}3x-5=0\\2x+2=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}3x=5\\2x=-2\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=3/5\\x=-1\end{array} \right.\)
    Vậy S={3/5;-1}

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

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