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\([\)6+(\(\dfrac{1}{2}\))3\(]\):\(\dfrac{3}{12}\)=\([\)6+\(\dfrac{1}{8}\)\(]\):\(\dfrac{1}{4}\)=\(\dfrac{49}{8}\):\(\dfrac{1}{4}\)=\(\dfrac{49}{2}\).
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
11 3/4-[65/4-41/2]+1-2/3
47/4-[65/4-41/2]+1-2/3
47/4-(-17/4)+1-2/3
16/1+1-2/3
16/1+1/1-2/3
17/1-2/3
51/3-2/3
49/3
707: [(2x -5) + 74] = 42 - 32
(2x -5) + 74 = 101
2x - 5 + 74 = 101
2x + 69 = 101
2x = 32 = 25
=> x = 5
707: [(2x -5) + 74] = 42 - 32
(2x -5) + 74 = 101
2x - 5 + 74 = 101
2x + 69 = 101
2x = 32 = 25
=> x = 5
\(50-\left[30-\left(9-4\right)^2\right]=50-\left(30-5^2\right)=50-\left(30-25\right)=50-5=45\)
\(\left[\frac{1}{33}+\frac{31}{333}-\left(\frac{341}{3333}\right)^2\right]\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
=\(\left[\frac{1}{33}+\frac{31}{333}-\left(\frac{341}{3333}\right)^2\right]\times\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
=\(\left[\frac{1}{33}+\frac{31}{333}-\left(\frac{341}{3333}\right)^2\right]\times\left(\frac{0}{6}\right)\)
=\(\left[\frac{1}{33}+\frac{31}{333}-\left(\frac{341}{3333}\right)^2\right]\times0\)
=0