Question

Tìm x biết:

a, x² - 16 = 0

b, (x - 3)² - x (x + 3) = 18

Answer 1

\(a,x^2-16=0\\ \Leftrightarrow\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ Vậy...\)

\(b,\left(x-3\right)^2-x\left(x+3\right)=18\\ \Leftrightarrow x^2-6x+9-x^2-3x=18\\ \Leftrightarrow-9x+9=18\\ \Leftrightarrow-9x=9\Leftrightarrow x=-1\\ vậy...\)

Answer 2

Lời giải:

\(a)\)

$x^2 - 16 = 0$

$\to x^2 - 4^2 = 0$

$\to (x - 4)(x + 4) = 0$

\(\rightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

Vậy ...

\(b)\)

$(x - 3)^2 - x(x + 3) = 18$

\(\rightarrow x^2-6x+9-x^2-3x=18\\ \rightarrow-9x=18-9\\ \rightarrow-9x=9\\ \rightarrow x=\dfrac{9}{-9}=-1\)

Vậy ...