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

Toán Lớp 8: a) $x^{3}$ – 16x = 0 b) $x^{4}$ – $2x^{3}$ + $10x^{2}$ – 20x = 0 c) 5x (2x + 3) + 7( 3 + 2x) = 0 d) (5 – 3x) . 4x – 15( 3x – 5)= 0

Tháng Bảy 16, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: a) $x^{3}$ – 16x = 0
b) $x^{4}$ – $2x^{3}$ + $10x^{2}$ – 20x = 0
c) 5x (2x + 3) + 7( 3 + 2x) = 0
d) (5 – 3x) . 4x – 15( 3x – 5)= 0

Comments ( 2 )

  1. trucoanh55
    trucoanh55
    16 Tháng Bảy, 2022 at 10:36 sáng
    Reply
    #tnvt
    x^3-16x=0
    <=>x(x^2-16)=0
    <=>x(x^2-4^2)=0
    <=>x(x-4)(x+4)=0
    <=>[(x=0),(x-4=0),(x+4=0):}
    <=>[(x=0),(x=4),(x=-4):}
    Vậy x\in{0;+-4}
    $\\$
    x^4-2x^3+10x^2-20x=0
    <=>x^3(x-2)+10x(x-2)=0
    <=>(x-2)(x^3+10x)=0
    <=>x(x-2)(x^2+10)=0
    <=>[(x=0),(x-2=0),(x^2+10=0):}
    <=>[(x=0),(x=2),(x^2=-10(\text{Vô lý})):}
    Vậy x\in{0;2}
    $\\$
    5x(2x+3)+7(3+2x)=0
    <=>5x(2x+3)+7(2x+3)=0
    <=>(2x+3)(5x+7)=0
    <=>$\left[\begin{matrix} x=-\dfrac{3}{2} \\ x=-\dfrac{7}{5}\end{matrix}\right.$
    Vậy x\in{-3/2;-7/5}
    $\\$
    (5-3x).4x-15.(3x-5)=0
    <=>4x.(5-3x)+15(5-3x)=0
    <=>(5-3x)(4x+15)=0
    <=>$\left[\begin{matrix} x=\dfrac{5}{3} \\ x=-\dfrac{15}{4}\end{matrix}\right.$
    Vậy x\in{5/3;-15/4}

  2. thucquyenlan
    thucquyenlan
    16 Tháng Bảy, 2022 at 10:36 sáng
    Reply
    Giải đáp:
     
    Lời giải và giải thích chi tiết:
     a, x^3 – 16x = 0
    ⇔ x(x^2 – 16) = 0
    ⇔ x(x – 4)(x + 4) = 0
    ⇒ $\left[\begin{matrix} x=0\\ x-4=0\\x + 4=0\end{matrix}\right.$ ⇒ $\left[\begin{matrix} x=0\\ x=4\\x = -4\end{matrix}\right.$
    Vậy x ∈ {-4;0;4}
    b, x^4 – 2x^3+ 10x^2 – 20x = 0
    ⇔ x^3 (x – 2) + 10x(x – 2) = 0
    ⇔ x(x – 2)(x^2 + 10) = 0
    ⇒ $\left[\begin{matrix} x=0\\ x-2=0\\x^2 + 10 =0\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x=0\\ x= 2\\x^2 = -10 (loại vì x^2 ≥0)\end{matrix}\right.$
    Vậy x = 0 hoặc x = 2
    c, 5x(2x + 3) + 7( 3+ 2x) = 0
    ⇔ (2x + 3)(5x + 7)= 0
    ⇒ $\left[\begin{matrix} 2x+3=0\\ 5x + 7=0\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} 2x=-3\\ 5x =-7\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x=\dfrac{-3}{2}\\ x =\dfrac{-7}{5}\end{matrix}\right.$
    Vậy x = -3/2 hoặc x = -7/5
    d, (5-3x).4x – 15(3x – 5) =0
    ⇔ (5 – 3x).4x + 15(5 – 3x) = 0
    ⇔(5 – 3x)(4x + 15) = 0
    ⇒ $\left[\begin{matrix} 5 – 3x = 0\\ 4x + 15 = 0\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} 3x = 5\\ 4x = -15\end{matrix}\right.$
    ⇒ $\left[\begin{matrix} x = \dfrac{5}{3}\\ x =\dfrac{-15}{4} \end{matrix}\right.$
    Vậy x = 5/3 hoặc x = -15/4

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

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