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

Toán Lớp 7: Tìm nghiệm của các đa thức sau: a) (x2 – 9)(x + l); b) x2 + 4x – 5; c) x2+ 9x + 20; d) x2 – x – 20; e) 2×2 +7x + 6; f) 3×2 + x –

Tháng Mười Hai 5, 2022 Toán học 2 Comments 0 views

Toán Lớp 7: Tìm nghiệm của các đa thức sau:
a) (x2 – 9)(x + l); b) x2 + 4x – 5;
c) x2+ 9x + 20; d) x2 – x – 20;
e) 2×2 +7x + 6; f) 3×2 + x – 4.

Comments ( 2 )

  1. chauhoangmy98
    chauhoangmy98
    5 Tháng Mười Hai, 2022 at 1:47 chiều
    Reply
    #Sad
    a)
    (x^2-9)(x+1) = 0
    ⇔ (x-3)(x+3)(x+1) = 0
    ⇔ \(\left[ \begin{array}{l}x-3=0\\x+3=0\\x+1=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=3\\x=-3\\x=-1\end{array} \right.\) 
    \text{Vậy S=} {3; -3; -1}
    b)
    x^2+4x-5 = 0
    ⇔ x^2+5x-x-5 = 0
    ⇔ (x^2+5x)-(x+5) = 0
    ⇔ x(x+5)-(x+5) = 0
    ⇔ (x-1)(x+5) = 0
    ⇔ \(\left[ \begin{array}{l}x-1=0\\x+5=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=1\\x=-5\end{array} \right.\) 
    \text{Vậy S=} {1; -5}
    c)
    x^2+9x+20 = 0
    ⇔ x^2+4x+5x+20 = 0
    ⇔ (x^2+4x)+(5x+20) = 0
    ⇔ x(x+4)+5(x+4) = 0
    ⇔ (x+5)(x+4)
    ⇔ \(\left[ \begin{array}{l}x+5=0\\x+4=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=-5\\x=-4\end{array} \right.\) 
    \text{Vậy S=} {-5; -4}
    d)
    x^2-x-20 = 0
    ⇔ x^2-5x+4x-20 = 0
    ⇔ (x^2-5x)+(4x-20) = 0
    ⇔ x(x-5)+4(x-5) = 0
    ⇔ (x+4)(x-5) = 0
    ⇔ \(\left[ \begin{array}{l}x+4=0\\x-5=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=-4\\x=5\end{array} \right.\) 
    \text{Vậy S=} {-4; 5}
    e)
    2x^2+7x+6 = 0
    ⇔ 2x^2+4x+3x+6 = 0
    ⇔ (2x^2+4x)+(3x+6) = 0
    ⇔ 2x(x+2)+3(x+2) = 0
    ⇔ (2x+3)(x+2) = 0
    ⇔ \(\left[ \begin{array}{l}2x+3=0\\x+2=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=-\dfrac{3}{2}\\x=-2\end{array} \right.\) 
    \text{Vậy S=} {-3/2; -2}
    f)
    3x^2+x-4 = 0
    ⇔ 3x^2+4x-3x-4 = 0
    ⇔ (3x^2-3x)+(4x-4) = 0
    ⇔ 3x(x-1)+4(x-1) = 0
    ⇔ (3x+4)(x-1) = 0
    ⇔ \(\left[ \begin{array}{l}3x+4=0\\x-1=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=-\dfrac{4}{3}\\x=1\end{array} \right.\) 
    \text{Vậy S=} {-4/3; 1}

  2. tuanh
    tuanh
    5 Tháng Mười Hai, 2022 at 1:46 chiều
    Reply
    ***Lời giải***
    a)
    (x^2-9)(x+1)=0
    <=>\(\left[ \begin{array}{l}x^2-9=0\\x+1=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x^2=9\\x=-1\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=3\\x=-3\\x=-1\end{array} \right.\)
    b)
    x^2+ 4x – 5=0
    <=>x^2-x+5x – 5=0
    <=>x(x-1)+5(x-1)=0
    <=>(x-1)(x+5)=0
    <=>\(\left[ \begin{array}{l}x-1=0\\x+5=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=1\\x=-5\end{array} \right.\) 
    c)
    x^2+ 9x + 20=0
    <=>x^2+ 5x+4x + 20=0
    <=>x(x+5)+4(x+5)=0
    <=>(x+4)(x+5)=0
    <=>\(\left[ \begin{array}{l}x+4=0\\x+5=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=-4\\x=-5\end{array} \right.\)
    d)
    x^2+4x-5x-20=0
    <=>x(x+4)-5(x+4)=0
    <=>(x+4)(x-5)=0
    <=>\(\left[ \begin{array}{l}x+4=0\\x-5=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=-4\\x=5\end{array} \right.\) 
    e)
    2x^2 +7x + 6=0
    <=>2x^2 +4x +3x + 6=0
    <=>2x(x+2)+3(x+2)=0
    <=>(x+2)(2x+3)=0
    <=>\(\left[ \begin{array}{l}x+2=0\\2x+3=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=-2\\2x=-3\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=-2\\x=\dfrac{-3}{2}\end{array} \right.\) 
    f)
    3x^2 + x – 4=0
    <=>3x^2 -3x+4x – 4=0
    <=>3x(x-1)+4(x-1)=0
    <=>(x-1)(3x+4)=0
    <=>\(\left[ \begin{array}{l}x-1=0\\3x+4=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=1\\3x=-4\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=1\\x=\dfrac{-4}{3}\end{array} \right.\) 

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

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