K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NL
2
a; 5\(x\) - 7 = 3\(x\) + 9
5\(x\) - 3\(x\) = 9 + 7
2\(x\) = 16
\(x\) = 16: 2
\(x\) = 8
Vậy \(x=8\)
b; 1\(\dfrac{3}{4}\)\(x\) + 1\(\dfrac{1}{2}\) = - \(\dfrac{4}{5}\)
\(\dfrac{7}{4}\)\(x\) + \(\dfrac{3}{2}\) = - \(\dfrac{4}{5}\)
\(\dfrac{7}{4}\)\(x\) = - \(\dfrac{4}{5}\) - \(\dfrac{3}{2}\)
\(\dfrac{7}{4}\)\(x\) = - \(\dfrac{23}{10}\)
\(x\) = - \(\dfrac{23}{10}\) : \(\dfrac{7}{4}\)
\(x\) = - \(\dfrac{46}{35}\)
Vậy \(x=-\dfrac{46}{35}\)
c; \(x\) + \(\dfrac{1}{2}\) = 25:23
\(x\) + \(\dfrac{1}{2}\) = 22
\(x\) + \(\dfrac{1}{2}\) = 4
\(x\) = 4 - \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{7}{2}\)
Vậy \(x=\dfrac{7}{2}\)
d; (\(x+\dfrac{1}{2}\))2 = \(\dfrac{4}{25}\)
\(\left[{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{2}{5}\\x+\dfrac{1}{2}=\dfrac{2}{5}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{2}{5}-\dfrac{1}{2}\\x=-\dfrac{2}{5}+\dfrac{1}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{9}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)
vậy \(x\) \(\in\) {- \(\dfrac{9}{10}\); - \(\dfrac{1}{10}\)}