Question

Bài 3: Tìm x biết

a) 4x^2 - 49 = 0;

b) x^2 + 36 = 12x;

c) 116x^2 - x + 4 = 0;

d) x^3 - 3 căn bậc 3x^2 + 9x - 3 căn bậc 3 = 0;

Answer 1

a) \(4x^2-49=0\)

<=> \(\left(2x-7\right)\left(2x+7\right)=0\)

<=> \(\left\{{}\begin{matrix}2x-7=0\\2x+7=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)

b) x² + 36 = 12x

<=>x² + 36 - 12x=0

<=> (x-6)²=0

<=> x-6 =0

<=> x=6

Answer 2

d) x³ -3\(\sqrt{3}\) x²+9x - 3\(\sqrt{3}\) =0

<=> \(\left(x-\sqrt{3}\right)^3=0\)

<=> \(x-\sqrt{3}=0\)

<=> \(x=\sqrt{3}\)