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|\(\frac32x\) + \(\frac12\)| = |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) \(\frac{x-3}{3}-1=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-3}{3}-\frac{3}{3}=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-6}{3}=\frac{x}{-4}\)
\(\Leftrightarrow-4\left(x-6\right)=3x\)
\(\Leftrightarrow-4x+24=3x\)
\(\Leftrightarrow24=3x+4x\)
\(\Leftrightarrow7x=24\)
\(\Leftrightarrow x=\frac{24}{7}\)
b) \(\frac{5}{8}-\left(x-\frac{1}{2}\right)=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}-x+\frac{1}{2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}+\frac{4}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow\frac{9}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{9}{8}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{15}{8}\)
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
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
a) 2012 x 8 - 8 x 2011
= 8 x (2012 - 2011)
= 8 x 1
= 8
b) \(\frac{7}{8}.2\frac{2}{3}-\frac{5}{8}.2\frac{2}{3}\)
\(=\frac{7}{8}.\frac{8}{3}-\frac{5}{8}.\frac{8}{3}\)
\(=\frac{8}{3}.\left(\frac{7}{8}-\frac{5}{8}\right)\)
\(=\frac{8}{3}.\frac{2}{8}\)
\(=\frac{16}{6}=\frac{8}{3}\)
1,
a,\(2012.8-8.2011\)
\(=8\left(2012-2011\right)\)
\(=8.1\)
\(=8\)
b.\(\frac{7}{8}.2\frac{2}{3}-\frac{5}{8}.2\frac{2}{3}\)
\(=2\frac{2}{3}.\left(\frac{7}{8}-\frac{5}{8}\right)\)
\(=\frac{8}{3}.\frac{1}{4}\)
\(=\frac{2}{3}\)
a, 2012.8-8.2011=8.(2012-2011)=8.1=8
b, \(\frac{7}{8}.2\frac{2}{3}-\frac{5}{8}.2\frac{2}{3}=\frac{8}{3}.\left(\frac{7}{8}-\frac{5}{8}\right)=\frac{8}{3}.\frac{1}{4}=\frac{2}{3}\)
\(2018.8-8.2011=8.\left(2018-2011\right)\)
\(=8.7=56\)
\(\frac{7}{8}.2\frac{2}{3}-\frac{5}{8}.2\frac{2}{3}=\frac{8}{3}\left(\frac{7}{8}-\frac{5}{8}\right)\)
\(=\frac{8}{3}.\frac{2}{8}=\frac{2}{3}\)
a) 2012 x 8 - 8 x 2011
= 8 x (2012 - 2011)
= 8,1
b) \(\frac{7}{8}\times2\frac{2}{3}-\frac{5}{8}\times2\frac{2}{3}\)
\(=2\frac{2}{3}\times\left(\frac{7}{8}-\frac{5}{8}\right)\)
\(=\frac{8}{3}\times\frac{1}{4}\)
\(=\frac{2}{3}\)
a) \(2012\times8-8\times2011\)
\(=\left(2012-2011\right)\times8\)
\(=1\times8\)
\(=8\)
b) \(\frac{7}{8}\times2\frac{2}{3}-\frac{5}{8}\times2\frac{2}{3}\)
\(=\left(\frac{7}{8}-\frac{5}{8}\right)\times2\frac{2}{3}\)
\(=\frac{2}{8}\times\frac{8}{3}\)\(=\frac{2}{3}\)
\(=\frac{1}{4}\times\frac{8}{3}\)