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

Toán Lớp 8: Tìm x: d, $x^{3}$ – $3x^{2}$ + x – 3 = 0 e, $9(x-1)^{2}$ = $4(2-3x)^{2}$

Toán Lớp 8: Tìm x:
d, $x^{3}$ – $3x^{2}$ + x – 3 = 0
e, $9(x-1)^{2}$ = $4(2-3x)^{2}$

Comments ( 2 )

  1. Giải đáp:
    d, x^3 – 3x^2 + x -3 = 0
    <=> x^2 ( x – 3) + ( x – 3) = 0
    <=> ( x^2 + 1)( x-3)=0
    <=> \(\left[ \begin{array}{l}x^2=-1 (KTM)\\x=3\end{array} \right.\) 
    Vậy S={ 3}
    e, 9( x -1)^2 = 4( 2 – 3x)^2
    <=>9x^2 – 18x + 9 = 36x^2 – 48x +16
    <=> 9x^2 – 36x^2 – 18x + 48x + 9 – 16 = 0
    <=> -27x^2 + 30x – 7= 0
    <=> 27x^2 – 30x + 7= 0
    <=> (x – 5/9)^2 = 4/81
    <=> \(\left[ \begin{array}{l}x-\frac{5}{9}=\frac{-2}{9}\\x- \frac{5}{9}= \frac{2}{9}\end{array} \right.\)
    <=> \(\left[ \begin{array}{l}x=\frac{1}{3}\\x=\frac{7}{9}\end{array} \right.\)
    Vậy S={ 1/3 ; 7/9}

  2. Giải đáp + Lời giải và giải thích chi tiết:
    d)
    x^{3}-3x^{2}+x-3=0
    <=>(x^{3}-3x^{2})+(x-3)=0
    <=>x^{2}(x-3)+(x-3)=0
    <=>(x-3)(x^{2}+1)=0
    Vì x^{2}+1>=1 với mọi x hay x^{2}+1\ne0 với mọi x
    =>x-3=0
    <=>x=3
    Vậy x=3
    e)
    9(x-1)^{2}=4(2-3x)^{2}
    <=>9(x^{2}-2x+1)=4(4-12x+9x^{2})
    <=>9x^{2}-18x+9=16-48x+36x^{2}
    <=>36x^{2}-9x^{2}+18x-48x+16-9=0
    <=>27x^{2}-30x+7=0
    <=>x^{2}-(10)/(9)x+(7)/(27)=0
    <=>(x-(5)/(9))^{2}-(4)/(81)=0
    <=>(x-(5)/(9))^{2}=(4)/(81)
    <=> \(\left[ \begin{array}{l}x-\dfrac{5}{9}=\dfrac{2}{9}\\x-\dfrac{5}{9}=-\dfrac{2}{9}\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=\dfrac{7}{9}\\x=\dfrac{1}{3}\end{array} \right.\) 
     

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

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