1 Tìm x bt a x^2-2x+1=0 b 9-6x+x^2=0 c 25x^2-10x+1=10 d 49x^2-28x+4 05/07/2021 Bởi Faith 1 Tìm x bt a x^2-2x+1=0 b 9-6x+x^2=0 c 25x^2-10x+1=10 d 49x^2-28x+4
$a)x^2-2x+1=0$ $⇔(x-1)^2=0$ $⇔x-1=0$ $⇔x=1$ $\text{Vậy $x=1$}$ $b)9-6x+x^2=0$ $⇔(3-x)^2=0$ $⇔3-x=0$ $⇔x=3$ $\text{Vậy $x=3$}$ $c)25x^2-10x+1=0$ $⇔(5x-1)^2=0$ $⇔5x-1=0$ $⇔5x=1$ $⇔x=\dfrac{1}{5}$ $\text{Vậy $x=\dfrac{1}{5}$}$ $d)49x^2-28x+4=0$ $⇔(7x-2)^2=0$ $⇔7x-2=0$ $⇔7x=2$ $⇔x=\dfrac{2}{7}$ $\text{Vậy $x=\dfrac{2}{7}$}$ Bình luận
$a)x^2-2x+1=0$
$⇔(x-1)^2=0$
$⇔x-1=0$
$⇔x=1$
$\text{Vậy $x=1$}$
$b)9-6x+x^2=0$
$⇔(3-x)^2=0$
$⇔3-x=0$
$⇔x=3$
$\text{Vậy $x=3$}$
$c)25x^2-10x+1=0$
$⇔(5x-1)^2=0$
$⇔5x-1=0$
$⇔5x=1$
$⇔x=\dfrac{1}{5}$
$\text{Vậy $x=\dfrac{1}{5}$}$
$d)49x^2-28x+4=0$
$⇔(7x-2)^2=0$
$⇔7x-2=0$
$⇔7x=2$
$⇔x=\dfrac{2}{7}$
$\text{Vậy $x=\dfrac{2}{7}$}$