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

Toán Lớp 8: Tìm x a. $x^{3} + $12x^{2} = 0 b. – $x^{2} + 2x + 15 = 0 c. 2x + $x^{2} = 35

Toán Lớp 8: Tìm x
a. $x^{3} + $12x^{2} = 0
b. – $x^{2} + 2x + 15 = 0
c. 2x + $x^{2} = 35

Comments ( 2 )

  1. Lời giải và giải thích chi tiết:
    a) x³ + 12x² = 0
    ⇔ x²( x + 12 ) = 0
    ⇔ \(\left[ \begin{array}{l}x²=0\\x+12=0\end{array} \right.\)
    ⇔ \(\left[ \begin{array}{l}x=0\\x=0-12\end{array} \right.\)
    ⇔ \(\left[ \begin{array}{l}x=0\\x=-12\end{array} \right.\) 
    Vậy x ∈ { 0 ; -12 }
    b) -x² + 2x + 15 = 0
    ⇔ -x² – 3x + 5x + 15 = 0
    ⇔ -x( x + 3 ) + 5( x + 3 ) = 0
    ⇔ -( x – 5 )( x + 3 ) = 0
    ⇔ \(\left[ \begin{array}{l}x-5=0\\x+3=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=5\\x=-3\end{array} \right.\) 
    Vậy x ∈ { 5 ; -3 }
    c) 2x + x² = 35
    ⇔ x² + 2x – 35 = 0
    ⇔ x² – 5x + 7x – 35 = 0
    ⇔ x( x – 5 ) + 7( x – 5 ) = 0
    ⇔ ( x + 7 )( x – 5 ) = 0
    ⇔ \(\left[ \begin{array}{l}x+7=0\\x-5=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=-7\\x=5\end{array} \right.\) 
    Vậy x ∈ { -7 ; 5 }
    CHÚC BẠN HỌC TỐT!

  2. a) x^3+12x^2=0
    <=> x^2(x+12)=0
    <=> [(x^2=0),(x+12=0):}<=>[(x=0),(x=-12):}
    Vậy S={0;-12}
    b) -x^2+2x+15=0
    <=> -x^2+5x-3x+15=0
    <=> -x(x-5)-3(x-5)=0
    <=> (x-5)(-x-3)=0
    <=> [(x-5=0),(-x-3=0):}<=>[(x=5),(x=-3):}
    Vậy S={5;-3}
    c) 2x+x^2=35
    <=> x^2+2x-35=0
    <=> x^2-5x+7x-35=0
    <=> x(x-5)+7(x-5)=0
    <=> (x-5)(x+7)=0
    <=> [(x-5=0),(x+7=0):}<=>[(x=5),(x=-7):}
    Vậy S={5;-7}

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