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

Toán Lớp 8: giúp mình với (┬┬﹏┬┬) bài 2: Tìm x a) 4x-16x^3=0 ; b) x^2-8x-16=0 c) x^3-3/2x^2+3/4x-1/8=0 d) x (x-5)+2(x-5)=0

Toán Lớp 8: giúp mình với (┬┬﹏┬┬)
bài 2: Tìm x
a) 4x-16x^3=0 ; b) x^2-8x-16=0
c) x^3-3/2x^2+3/4x-1/8=0
d) x (x-5)+2(x-5)=0

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết:
    a) 4x-16x^3=0
    <=>4x(1-4x^2)=0
    <=>4x(1-2x)(1+2x)=0
    <=>\(\left[ \begin{array}{l}4x=0\\1-2x=0\\1+2x=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.\)
    Vậy S={0;1/2;-1/2}
    b) x^2-8x+16=0 ( sửa đề)
    <=> (x-4)^2=0
    <=>x-4=0
    <=>x=4
    Vậy S={4}
    c) x^3 – 3/2x^2 + 3/4x – 1/8 = 0
    <=> x^3-3*x^2*1/2+3*x*(1/2)^2-(1/2)^3=0
    <=>(x-1/2)^3=0
    <=>x-1/2=0
    <=>x=1/2
    Vậy S={1/2}
    d) x(x-5)+2(x-5)=0<=>(x-5)(x+2)=0
    <=>\(\left[ \begin{array}{l}x-5=0\\x+2=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)
    Vậy S={5;-2}. 

  2. a)4x-16x³=0
    ⇔4x(1-4x²)=0
    ⇔4x[1²-(2x)²]=0
    ⇔4x(1+2x)(1-2x)=0
    ⇔\(\left[ \begin{array}{l}4x=0\\1+2x=0\\1-2x=0\end{array} \right.\)
    ⇔\(\left[ \begin{array}{l}x=0\\x=-\dfrac{1}{2}\\x=\dfrac{1}{2}\end{array} \right.\)
    Vậy S={0;-1/2;1/2}
    b) Sửa đề:x²-8x-16=0
    →x²-8x+16=0
    x²-8x+16=0
    ⇔x²-2.x.4+4²=0
    ⇔(x-4)²=0
    ⇔x-4=0
    ⇔x=4
    Vậy S={4}
    c)x³-3/2x²+3/4x-1/8=0
    ⇔x³-3.x². 1/2+3.x. (1/2)^2-(1/2)^3=0
    ⇔(x-1/2)^3=0
    ⇔x-1/2=0
    ⇔x=1/2
    Vậy S={1/2}
    d)x(x-5)+2(x-5)=0
    ⇔(x-5)(x+2)=0
    ⇔\(\left[ \begin{array}{l}x-5=0\\x+2=0\end{array} \right.\)
    ⇔\(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)
     Vậy S={5;-2}

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

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