Question

Tìm x

a, x\(^2\)+12x+36=0

b, x\(^2\)-1=0 

c, 25x\(^2\)-9=0

Answer 1

a: \(x^2+12x+36=0\)

\(\Leftrightarrow\left(x+6\right)^2=0\)

\(\Leftrightarrow x+6=0\)

hay x=-6

b: Ta có: \(x^2-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c: Ta có: \(25x^2-9=0\)

\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

Answer 2

Lời giải:
a. $x^2+12x+36=0$

$\Leftrightarrow (x+6)^2=0$

$\Leftrightarrow x+6=0$

$\Leftrightarrow x=-6$

b.

$x^2-1=0$

$\Leftrightarrow (x-1)(x+1)=0$
$\Leftrightarrow x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=1$ hoặc $x=-1$

c. 

$25x^2-9=0$

$\Leftrightarrow (5x)^2-3^2=0$

$\Leftrightarrow (5x-3)(5x+3)=0$

$\Leftrightarrow 5x-3=0$ hoặc $5x+3=0$

$\Leftrightarrow x=\frac{3}{5}$ hoặc $x=-\frac{3}{5}$