Câu hỏi của Mộc Lung Hoa

Question

3. Tìm x, biết:

a) x3 - 1/9 = 0

b) 2x - 2y - x2 + 2xy - y2 = 0

c) x(x -30 = x - 3 = 0

d) x2 ( x - 3) + 27 - 9x = 0

lê thị hương giang · Accepted Answer

a,$x^3-\dfrac{1}{9}=0$

$\Rightarrow x^3-\left(\dfrac{1}{3}\right)^3=0$

$\Rightarrow\left(x-\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)=0$

$\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=0\x^2+\dfrac{1}{3}x+\dfrac{1}{9}=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\x^2+\dfrac{1}{3}x=-\dfrac{1}{9}\end{matrix}\right.$

$\Rightarrow x=\dfrac{1}{3}$Câu trả lời: a,$x^3-\dfrac{1}{9}=0$

$\Rightarrow x^3-\left(\dfrac{1}{3}\right)^3=0$

$\Rightarrow\left(x-\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)=0$

$\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=0\x^2+\dfrac{1}{3}x+\dfrac{1}{9}=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\x^2+\dfrac{1}{3}x=-\dfrac{1}{9}\end{matrix}\right.$

$\Rightarrow x=\dfrac{1}{3}$