Bài 1:tìm x,biết a, 130-|x-10|=50 b,|2x-15|-45=0 c,(2x-5) ² =121 02/11/2021 Bởi Liliana Bài 1:tìm x,biết a, 130-|x-10|=50 b,|2x-15|-45=0 c,(2x-5) ² =121
Đáp án: a) \(\left[ \begin{array}{l}x=90\\x=-70\end{array} \right.\) b) \(\left[ \begin{array}{l}x=30\\x=-15\end{array} \right.\) c) \(\left[ \begin{array}{l}x=8\\x=-3\end{array} \right.\) Giải thích các bước giải: a) $130 – |x-10|=50$ $|x-10| = 80 $x – 10 = ±80$ ⇒ \(\left[ \begin{array}{l}x=90\\x=-70\end{array} \right.\) b,$|2x-15|-45=0$ $|2x-15|=45$ $2x – 15 = ± 45$ \(\left[ \begin{array}{l}x=30\\x=-15\end{array} \right.\) c) $(2x-5) ² =121$ $(2x – 5)² = 11²$ ⇒ $x = 8$ Bình luận
Bài làm : a, `130-|x-10|=50` `⇔|x-10|=80` `⇔` \(\left[ \begin{array}{l}x-10=80\\x-10=-80\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=90\\x=-70\end{array} \right.\) Vậy `x∈{90;-70}` b, `|2x-15|-45=0` `⇔|2x-15|=45` `⇔` \(\left[ \begin{array}{l}2x-15=45\\2x-15=-45\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}2x=60\\2x=-30\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=30\\x=-15\end{array} \right.\) Vậy `x∈{30 ; -15}` c, `(2x-5)² =121` `⇔ (2x-5)² = 11²` `⇔ 2x – 5 = 11` `⇔ 2x = 16` `⇔ x = 8` Vậy `x = 8` Bình luận
Đáp án:
a) \(\left[ \begin{array}{l}x=90\\x=-70\end{array} \right.\)
b) \(\left[ \begin{array}{l}x=30\\x=-15\end{array} \right.\)
c) \(\left[ \begin{array}{l}x=8\\x=-3\end{array} \right.\)
Giải thích các bước giải:
a) $130 – |x-10|=50$
$|x-10| = 80
$x – 10 = ±80$
⇒ \(\left[ \begin{array}{l}x=90\\x=-70\end{array} \right.\)
b,$|2x-15|-45=0$
$|2x-15|=45$
$2x – 15 = ± 45$
\(\left[ \begin{array}{l}x=30\\x=-15\end{array} \right.\)
c) $(2x-5) ² =121$
$(2x – 5)² = 11²$
⇒ $x = 8$
Bài làm :
a, `130-|x-10|=50`
`⇔|x-10|=80`
`⇔` \(\left[ \begin{array}{l}x-10=80\\x-10=-80\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=90\\x=-70\end{array} \right.\)
Vậy `x∈{90;-70}`
b, `|2x-15|-45=0`
`⇔|2x-15|=45`
`⇔` \(\left[ \begin{array}{l}2x-15=45\\2x-15=-45\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x=60\\2x=-30\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=30\\x=-15\end{array} \right.\)
Vậy `x∈{30 ; -15}`
c, `(2x-5)² =121`
`⇔ (2x-5)² = 11²`
`⇔ 2x – 5 = 11`
`⇔ 2x = 16`
`⇔ x = 8`
Vậy `x = 8`