Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)
\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)
\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)
=>-5/3x=55/212
hay x=-33/212
c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
=>x+3=19
hay x=16
a) 2 - ( \(5\frac{3}{8}\)x X - \(\frac{5}{24}\)) = \(\frac{5}{12}\)
\(5\frac{3}{8}\)x X - \(\frac{5}{24}\)= \(\frac{19}{12}\)
\(5\frac{3}{8}\)x X = \(\frac{43}{24}\)
X = \(\frac{1}{3}\)
b) \(1\frac{2}{9}\): ( \(3\frac{1}{3}\)x X + \(\frac{1}{6}\)) = \(\frac{22}{23}\)
\(3\frac{1}{3}\)x X + \(\frac{1}{6}\) = \(\frac{23}{18}\)
\(3\frac{1}{3}\)x X = \(\frac{10}{9}\)
X =\(\frac{1}{3}\)
C) \(\frac{4}{5}\)x X - \(\frac{1}{2}\)x X + \(\frac{3}{4}\)x X = \(\frac{7}{40}\)
( \(\frac{4}{5}-\frac{1}{2}+\frac{3}{4}\)) x X = \(\frac{7}{40}\)
\(\frac{21}{20}\) x X = \(\frac{7}{40}\)
X =\(\frac{1}{6}\)
|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|
\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)
\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)
\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)
Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}
|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0
|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|
\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)
Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
a/ \(\frac{6}{7}x=\frac{18}{23}\)
\(x=\frac{18}{23}:\frac{6}{7}=\frac{21}{23}\)
b/ \(2\frac{1}{2}x=\frac{5}{6}\)
\(=>\frac{5}{2}x=\frac{5}{6}\)
\(x=\frac{5}{6}:\frac{5}{2}=\frac{1}{3}\)
c/\(x:2\frac{3}{4}=9\frac{5}{8}\)
\(x:\frac{11}{4}=\frac{77}{8}\)
\(x=\frac{77}{8}\cdot\frac{11}{4}=\frac{847}{32}\)
d/\(7\frac{1}{7}\cdot\frac{1}{7}\cdot x=22\frac{1}{8}\)
\(\frac{50}{49}x=\frac{177}{8}\)
\(x=\frac{177}{8}:\frac{50}{49}=\frac{8673}{400}\)
\(a,\frac{6}{7}.x=\frac{18}{23}\) \(\Rightarrow x=\frac{18}{23}:\frac{6}{7}=\frac{18}{23}.\frac{7}{6}=\frac{21}{23}\)
\(b,2\frac{1}{2}.x=\frac{5}{6}\Rightarrow\frac{5}{2}.x=\frac{5}{6}\Rightarrow x=\frac{5}{6}:\frac{5}{2}=\frac{5}{6}.\frac{2}{5}=\frac{1}{3}\)
\(c,x:2\frac{3}{4}=9\frac{5}{8}\Rightarrow x:\frac{11}{4}=\frac{77}{8}\Rightarrow x=\frac{77}{8}.\frac{11}{4}=\frac{847}{32}\)
\(d,7\frac{1}{7}.\frac{1}{7}.x=22\frac{1}{8}\Rightarrow\frac{50}{49}.x=\frac{177}{8}\Rightarrow x=\frac{177}{8}:\frac{50}{49}=\frac{177}{8}.\frac{49}{50}=\frac{8673}{400}\)
b) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
=> 3.(x + 2) = (x - 5). (-4)
= 3x + 6 = -4x + 20
= 3x - 4x = 26
=> -x = 26
=> x = -26
c) \(\frac{x}{-2}=\frac{-8}{x}\)
=> x.x = -2 . (-8)
=> x2 = 16
=> x = 4
t i c k nha!!! 645646677778879078452352543546456457567564546567

\(\frac{5}{8}\)X-\(\frac{1}{8}\)=\(\frac{23}{4}\)
\(\frac{5}{8}\)X=\(\frac{23}{4}\)+\(\frac{1}{8}\)
\(\frac{5}{8}\)X=\(\frac{47}{8}\)
x=\(\frac{47}{8}\):\(\frac{5}{8}\)
X=\(\frac{47}{5}\)