giúp mk vs :<< 1/ |2x - 5| = x + 1 2/ |x - 1| = 2x + 1 09/11/2021 Bởi Kaylee giúp mk vs :<< 1/ |2x - 5| = x + 1 2/ |x - 1| = 2x + 1
Đáp án: Giải thích các bước giải: `1,|2x-5|=x+1` `⇒`\(\left[ \begin{array}{l}2x-5=x+1\\2x-5=-x-1\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x=6\\3x=4\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x=6\\x=\dfrac{4}{3}\end{array} \right.\) `2,|x-1|=2x+1` `⇒`\(\left[ \begin{array}{l}x-1=2x+1\\x-1=-2x-1\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x=-2\\x=0\end{array} \right.\) Bình luận
Đáp án: Giải thích các bước giải: 1/ |2x – 5| = x + 1 ⇔\(\left[ \begin{array}{l}2x-5=x+1\\2x-5=-x-1\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=6\\3x=4\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=6\\x=3/4\end{array} \right.\) Vậy x ∈ {6;3/4} 2/ |x – 1| = 2x + 1 ⇔\(\left[ \begin{array}{l}x-1=2x+1\\x-1=-2x-1\end{array} \right.\) ⇔\(\left[ \begin{array}{l}-x=2\\3x=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-2\\x=0\end{array} \right.\) Vậy x ∈ {-2;0} Bình luận
Đáp án:
Giải thích các bước giải:
`1,|2x-5|=x+1`
`⇒`\(\left[ \begin{array}{l}2x-5=x+1\\2x-5=-x-1\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=6\\3x=4\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=6\\x=\dfrac{4}{3}\end{array} \right.\)
`2,|x-1|=2x+1`
`⇒`\(\left[ \begin{array}{l}x-1=2x+1\\x-1=-2x-1\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=-2\\x=0\end{array} \right.\)
Đáp án:
Giải thích các bước giải:
1/ |2x – 5| = x + 1
⇔\(\left[ \begin{array}{l}2x-5=x+1\\2x-5=-x-1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=6\\3x=4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=6\\x=3/4\end{array} \right.\)
Vậy x ∈ {6;3/4}
2/ |x – 1| = 2x + 1
⇔\(\left[ \begin{array}{l}x-1=2x+1\\x-1=-2x-1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}-x=2\\3x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=0\end{array} \right.\)
Vậy x ∈ {-2;0}