Ta có: $(x+\dfrac{1}{2})^2 = \dfrac{1}{25}$ $⇒(x+\dfrac{1}{2})^2 = (\dfrac{1}{5})^2$ hoặc $(x+\dfrac{1}{2})^2 = (-\dfrac{1}{5})^2$ $⇒x+\dfrac{1}{2}=\dfrac{1}{5}$ hoặc $⇒x+\dfrac{1}{2}=-\dfrac{1}{5}$ $⇒x=-\dfrac{3}{10}$ hoặc $x=\dfrac{-7}{10}$ Vậy $x=-\dfrac{3}{10}$ hoặc $x=\dfrac{-7}{10}$ thì $(x+\dfrac{1}{2})^2 = \dfrac{1}{25}$. Bình luận
( x + $\frac{1}{2}$ )² = $\frac{1}{25}$ ⇒ ( x + $\frac{1}{2}$ )² = ( $\frac{1}{5}$ )² ⇒ x + $\frac{1}{2}$ = $\frac{1}{5}$ ⇒ x = $\frac{1}{5}$ – $\frac{1}{2}$ ⇒ x = $\frac{2}{10}$ – $\frac{5}{10}$ ⇒ x = $\frac{-3}{10}$ Bình luận
Ta có:
$(x+\dfrac{1}{2})^2 = \dfrac{1}{25}$
$⇒(x+\dfrac{1}{2})^2 = (\dfrac{1}{5})^2$ hoặc $(x+\dfrac{1}{2})^2 = (-\dfrac{1}{5})^2$
$⇒x+\dfrac{1}{2}=\dfrac{1}{5}$ hoặc $⇒x+\dfrac{1}{2}=-\dfrac{1}{5}$
$⇒x=-\dfrac{3}{10}$ hoặc $x=\dfrac{-7}{10}$
Vậy $x=-\dfrac{3}{10}$ hoặc $x=\dfrac{-7}{10}$ thì $(x+\dfrac{1}{2})^2 = \dfrac{1}{25}$.
( x + $\frac{1}{2}$ )² = $\frac{1}{25}$
⇒ ( x + $\frac{1}{2}$ )² = ( $\frac{1}{5}$ )²
⇒ x + $\frac{1}{2}$ = $\frac{1}{5}$
⇒ x = $\frac{1}{5}$ – $\frac{1}{2}$
⇒ x = $\frac{2}{10}$ – $\frac{5}{10}$
⇒ x = $\frac{-3}{10}$