a) (x^2-1)(2y+4)=0 b) /2x-1/-3=0 Trg đề mik chỉ còn 2 câu này là mik chx bt chứ mấy câu kia mik gửi bên kia r 12/07/2021 Bởi Rose a) (x^2-1)(2y+4)=0 b) /2x-1/-3=0 Trg đề mik chỉ còn 2 câu này là mik chx bt chứ mấy câu kia mik gửi bên kia r
a) $(x²-1).(2y-4)=0$ ⇒\(\left[ \begin{array}{l}x²-1=0\\2y-4=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x²=1\\2y=4\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=1 hoặc x=-1\\y=2\end{array} \right.\) Vậy $x∈${$1;-1$} ;$ y=2$ b) $|2x-1|-3=0$ $|2x-1|=3$ ⇒\(\left[ \begin{array}{l}2x-1=3\\2x-1=-3\end{array} \right.\) ⇒\(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\) Vậy $x∈${$2;-1$} Bình luận
Đáp án: a, `(x² – 1)(2y + 4) = 0` ⇒ \(\left[ \begin{array}{l}x² – 1 = 0\\2y + 4 =0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x²=1\\2y = -4\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x= ±1\\y = -2\end{array} \right.\) b, `| 2x – 1| – 3 =0` `| 2x – 1| = 0 + 3 = 3` ⇒\(\left[ \begin{array}{l}2x – 1 = 3\\2x – 1 = -3\end{array} \right.\) ⇒\(\left[ \begin{array}{l}2x = 4\\2x = -2\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\) `text{ @toanisthebest}` Bình luận
a) $(x²-1).(2y-4)=0$
⇒\(\left[ \begin{array}{l}x²-1=0\\2y-4=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x²=1\\2y=4\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=1 hoặc x=-1\\y=2\end{array} \right.\)
Vậy $x∈${$1;-1$} ;$ y=2$
b) $|2x-1|-3=0$
$|2x-1|=3$
⇒\(\left[ \begin{array}{l}2x-1=3\\2x-1=-3\end{array} \right.\)
⇒\(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Vậy $x∈${$2;-1$}
Đáp án:
a, `(x² – 1)(2y + 4) = 0`
⇒ \(\left[ \begin{array}{l}x² – 1 = 0\\2y + 4 =0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x²=1\\2y = -4\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x= ±1\\y = -2\end{array} \right.\)
b, `| 2x – 1| – 3 =0`
`| 2x – 1| = 0 + 3 = 3`
⇒\(\left[ \begin{array}{l}2x – 1 = 3\\2x – 1 = -3\end{array} \right.\)
⇒\(\left[ \begin{array}{l}2x = 4\\2x = -2\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
`text{ @toanisthebest}`