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

Toán Lớp 8: Tìm x biết : 1. 7x(x-2) = x-2 2. 8x^3 – 12x^2 + 6x – 1 = 0 3. x^3 – 25x = 0 4. x^3 + 6x^2 + 9x = 0 5. x^2 + 9x + 20 = 0

Toán Lớp 8: Tìm x biết :
1. 7x(x-2) = x-2
2. 8x^3 – 12x^2 + 6x – 1 = 0
3. x^3 – 25x = 0
4. x^3 + 6x^2 + 9x = 0
5. x^2 + 9x + 20 = 0

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết:
    1, 7x(x-2) = x-2
    <=>7x(x-2)+ (x-2 )= 0
    <=>(x-2)(7x+1)=0
    =>\(\left[ \begin{array}{l}x-2=0\\7x+1=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=2\\x= \dfrac{-1}{7}\end{array} \right.\)
    2,8x^3 – 12x^2 + 6x – 1 = 0
    <=> ( 2x)^(3)  – 3.(2x)^(2).1 + 3.2x.1 – 1^3 = 0
    <=> ( 2x – 1)^3 = 0
    => 2x – 1 =0
    <=> x = 1/2
    3, x^3 – 25x = 0
    <=> x( x^2 – 25) = 0
    <=> x( x – 5)(x + 5) = 0
    =>\(\left[ \begin{array}{l}x=0\\x-5=0\\x+5=0\end{array} \right.\)
    =>\(\left[ \begin{array}{l}x=0\\x=5\\x=-5\end{array} \right.\)
    4,x^3 + 6x^2 + 9x = 0
    ⇔ x(x^2+6x+9)=0
    ⇔ x(x+3)^2=0
    =>\(\left[ \begin{array}{l}x=0\\x+3=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.\)
    5,x^2 + 9x + 20 = 0
    <=>x^2+4x+5x+20=0
    <=>x(x+4)+5(x+4)=0
    <=>(x+4)(x+5)=0
     =>\(\left[ \begin{array}{l}x+4=0\\x+5=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=-4\\x=-5\end{array} \right.\)

  2. Giải đáp + Lời giải và giải thích chi tiết:
    a)$7x(x-2)=x-2$
    $→7x(x-2)-(x-2)=0$
    $→(7x-1)(x-2)=0$
    →\(\left[ \begin{array}{l}x-2=0\\7x-1=0\end{array} \right.\)
    →\(\left[ \begin{array}{l}x=2\\7x=1\end{array} \right.\)
    →\(\left[ \begin{array}{l}x=2\\x=\frac{1}{7} \end{array} \right.\)
    Vậy x=2 ; x=$\frac{1}{7}$ 
    b)$8x^{3}-12x^{2}+6x-1=0$
    $→(2x-1)^{3}=0=0^{3}$
    $→2x-1=0$ $→2x=1$
    $→x=\frac{1}{2}$ 
    Vậy x=$\frac{1}{2}$
    c)$x^{3}-25x=0$
    $→x(x^{2}-25)=0$
    $→x(x-5)(x+5)=0$
    $→x=0$   hoặc   $x-5=0$   hoặc   $x+5=0$ $
    →x=0$     $x=5$      $x=-5$
    Vậy x∈{-5;0;5}
    d)$x^{3}+6x^{2}+9x=0$
    $→x(x^{2}+6x+9)=0$
    $→x(x+3)^{2}=0$
    →\(\left[ \begin{array}{l}x=0\\(x+3)^{2}=0=0^2\end{array} \right.\)
    →\(\left[ \begin{array}{l}x=0\\x+3=0\end{array} \right.\)
    → \(\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.\) 
    Vậy x=0 ; x=-3
    e)$x^{2}+9x+20=0$
    $→x^2+4x+5x+20=0$
    $→(x^2+4)+(5x+20)=0$
    $→x(x+4)+5(x+4)=0$ $→(x+5)(x+4)=0$
    →\(\left[ \begin{array}{l}x+5=0\\x+4=0\end{array} \right.\)
    →\(\left[ \begin{array}{l}x=-5\\x=-4\end{array} \right.\) 
    Vậy x=-5 ; x=-4

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

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