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|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)
|7 - \(\dfrac{3}{4}x\)| - \(\dfrac{3}{2}\) = 2
|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)
|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)
\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)
5 - |\(x-3\)| = 5
|\(x-3\)| = 5 - 5
|\(x-3\)| = 0
\(x-3\) = 0
\(x\) = 3
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a) \(\left(6-\frac{2}{3}+\frac{1}{2}\right)\)- \(\left(5+\frac{5}{3}-\frac{2}{3}\right)\)- \(\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
= \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{2}{3}-3+\frac{7}{3}-\frac{5}{2}\)
= \(-2+\frac{2}{3}-\frac{4}{2}\)
= \(\frac{-12}{6}+\frac{4}{6}-\frac{12}{6}\)= \(\frac{4}{6}=\frac{2}{3}\)
a/\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{2}{3}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)..\)
\(=\left(6-\frac{1}{6}\right)-\left(5+1\right)-\left(3-\frac{-1}{6}\right)\)
\(=6-\frac{1}{6}-\left(5+1\right)-3-\frac{1}{6}\)
\(=\left[6-\left(5+1\right)\right]-\frac{1}{6}-\frac{1}{6}-3\)
\(=0-\frac{1}{6}-\frac{1}{6}-3\)
\(=\frac{-1}{6}-\frac{1}{6}-3\)
\(=\frac{-10}{3}\)
a, \(\frac{-3}{7}+\frac{5}{13}-\frac{4}{7}+\frac{8}{13}\)
\(=\frac{-3}{7}-\frac{4}{7}+\frac{5}{13}+\frac{8}{13}\)
\(=-\frac{7}{7}+\frac{13}{13}=-1+1=0\)
b, \(\frac{-5}{14}-\frac{2}{-14}+\frac{1}{8}+\frac{1}{8}\)
\(=\frac{-5}{14}+\frac{2}{14}+\frac{1}{8}+\frac{1}{8}\)
\(=-\frac{3}{14}+\frac{1}{4}=\frac{1}{28}\)
c,\(-\frac{5}{13}-\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)\)
\(=-\frac{5}{13}-\frac{3}{13}-\frac{3}{5}+\frac{4}{10}\)
\(=-\frac{8}{13}-\frac{3}{5}+\frac{4}{10}=-\frac{79}{65}+\frac{4}{10}=-\frac{53}{65}\)
d, \(\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{4}{6}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{2}{3}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{1}{2}-\frac{9}{7}-\frac{12}{7}+\frac{2}{3}\right)+\frac{5}{9}\right]\)
\(=-\frac{65}{24}+\frac{5}{9}=-2\frac{11}{72}\)
B={[5^10. 7^3 - 25^2 . 49^2] : [(125 . 7)^3+5^9 . 14^3]} - {[2^12-4^6 . 9^2] : [(2^2.3)^6+8^4 . 3^5]
Sửa đề: \(5^9\cdot49^2\)
\(=\dfrac{5^{10}\cdot7^3-5^9\cdot7^4}{5^9\cdot7^3+5^9\cdot14^3}-\dfrac{2^{12}-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)
\(=\dfrac{5^9\cdot7^3\left(5-7\right)}{5^9\cdot7^3\left(1+8\right)}-\dfrac{2^{12}\left(1-3^4\right)}{2^{12}\left(3^6+3^5\right)}=\dfrac{-2}{9}+\dfrac{80}{972}\)
=-34/243
3: \(=\dfrac{13}{5}\left(-\dfrac{3}{14}+\dfrac{2}{5}+\dfrac{-11}{14}+\dfrac{3}{5}\right)\)
=0
2\(\dfrac{4}{5}\) + (- \(\dfrac{3}{7}\) + \(\dfrac{2}{3}\)) : - \(\dfrac{5}{14}\)
= \(\dfrac{14}{5}\) + (\(\dfrac{-9}{21}\) + \(\dfrac{14}{21}\)) : - \(\dfrac{5}{14}\)
= \(\dfrac{14}{5}\) + \(\dfrac{5}{21}\) x (- \(\dfrac{14}{5}\))
= \(\dfrac{14}{5}\) - \(\dfrac{2}{3}\)
= \(\dfrac{42}{15}\) - \(\dfrac{10}{15}\)
= \(\dfrac{32}{15}\)