Giải phương trình: a) 5y ² + y = 0 b) 5/y-3 + 4/y+3 = y-5/y ²-9 11/10/2021 Bởi Eva Giải phương trình: a) 5y ² + y = 0 b) 5/y-3 + 4/y+3 = y-5/y ²-9
a) $5y^2+y=0$ $⇔y(5y+1)=0$ \(⇔\left[ \begin{array}{l}y=0\\5y+1=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}y=0\\y=\dfrac{-1}{5}\end{array} \right.\) Vậy $S=\bigg\{0;\dfrac{-1}{5}\bigg\}$ b) $\dfrac{5}{y-3}+\dfrac{4}{y+3}=\dfrac{y-5}{y^2-9}$ ĐK: $x\neq\pm3$ $⇔\dfrac{5}{y-3}+\dfrac{4}{y+3}=\dfrac{y-5}{(y-3)(y+3)}$ $⇔\dfrac{5(y+3)}{(y-3)(y+3)}+\dfrac{4(y-3)}{(y-3)(y+3)}=\dfrac{y-5}{(y-3)(y+3)}$ $⇒5y+15+4y-12=y-5$ $⇔8y=-8$ $⇔y=-1(tm)$ Vậy $S=\{-1\}$ Bình luận
Đáp án: Giải thích các bước giải: a,$5y^{2}$+y=0 ⇔y(5y+1)=0 ⇔\(\left[ \begin{array}{l}y=0\\5y+1=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}y=0\\y=(-1)/5\end{array} \right.\) Vậy … Bình luận
a) $5y^2+y=0$
$⇔y(5y+1)=0$
\(⇔\left[ \begin{array}{l}y=0\\5y+1=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}y=0\\y=\dfrac{-1}{5}\end{array} \right.\)
Vậy $S=\bigg\{0;\dfrac{-1}{5}\bigg\}$
b) $\dfrac{5}{y-3}+\dfrac{4}{y+3}=\dfrac{y-5}{y^2-9}$ ĐK: $x\neq\pm3$
$⇔\dfrac{5}{y-3}+\dfrac{4}{y+3}=\dfrac{y-5}{(y-3)(y+3)}$
$⇔\dfrac{5(y+3)}{(y-3)(y+3)}+\dfrac{4(y-3)}{(y-3)(y+3)}=\dfrac{y-5}{(y-3)(y+3)}$
$⇒5y+15+4y-12=y-5$
$⇔8y=-8$
$⇔y=-1(tm)$
Vậy $S=\{-1\}$
Đáp án:
Giải thích các bước giải:
a,$5y^{2}$+y=0
⇔y(5y+1)=0
⇔\(\left[ \begin{array}{l}y=0\\5y+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}y=0\\y=(-1)/5\end{array} \right.\)
Vậy …