d) Ta có: \(\left(y+3\right)^2\ge0\forall y\)
\(\left(y+5\right)^2\ge0\forall y\)
Do đó: \(\left(y+3\right)^2+\left(y+5\right)^2\ge0\forall y\)
mà \(\left(y+3\right)^2+\left(y+5\right)^2=0\)
nên \(\left\{{}\begin{matrix}\left(y+3\right)^2=0\\\left(y+5\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+3=0\\y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\y=-5\end{matrix}\right.\)
Vậy: y=-3 và y=-5
a, y2 + 7y + 2 = 0
⇔ y(y + 7) + 2 = 0
⇔ y = \(-\frac{7-\sqrt{41}}{2}\) ; \(-\frac{7+\sqrt{41}}{2}\)
*(Nâng cao): - 0.29843788... ; - 6.70156211...
Vậy y = \(-\frac{7-\sqrt{41}}{2}\) ; \(-\frac{7+\sqrt{41}}{2}\).
Chúc bạn học tốt!
