tìm x biết rằng: a) x-1+2(x-1)^2=0 b) (x+3)(x^2-3x+9)-x(x^2-3)=18 c) (x-4)(x^2+4x+16)-x(x^2-6)=2 d)3x^2+3=10x

tìm x biết rằng:
a) x-1+2(x-1)^2=0
b) (x+3)(x^2-3x+9)-x(x^2-3)=18
c) (x-4)(x^2+4x+16)-x(x^2-6)=2
d)3x^2+3=10x

1 bình luận về “tìm x biết rằng: a) x-1+2(x-1)^2=0 b) (x+3)(x^2-3x+9)-x(x^2-3)=18 c) (x-4)(x^2+4x+16)-x(x^2-6)=2 d)3x^2+3=10x”

  1. a) x-1+2(x-1)^2=0
    (x-1)[1 + 2(x-1)] = 0
    (x-1)(2x – 1) = 0
    => \(\left[ \begin{array}{l}x-1=0\\2x-1=0\end{array} \right.\)
    => \(\left[ \begin{array}{l}x=1\\x=\frac{1}{2} \end{array} \right.\) \frac{1}{2}
    Vậy…
    b) (x+3)(x^2-3x+9)-x(x^2-3)=18
    x^3 + 27 – x^3 + 3x – 18 = 0
    3x + 9 = 0
    3x = -9
    x = -3
    Vậy…
    c) (x-4)(x^2+4x+16)-x(x^2-6)=2
    x^3 – 64 – x^3 + 6x – 2 = 0
    6x – 66 = 0
    6x = 66
    x = 11
    Vậy…
    d)3x^2+3=10x
    3x^2 – 10x + 3 = 0
    3x^2 – 9x – x + 3 = 0
    3x(x-3) – (x-3) = 0
    (x-3)(3x-1) = 0
    => \(\left[ \begin{array}{l}x-3=0\\3x-1=0\end{array} \right.\)
    => \(\left[ \begin{array}{l}x=3\\x=\frac{1}{3}\end{array} \right.\) 
    Vậy…

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