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A = \(\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}\)
A = \(\left(-\frac{1}{4}\right)+\frac{5}{6}-\frac{7}{12}\)
A = \(\frac{7}{12}-\frac{7}{12}\)
A = \(0\).
Mình làm câu A thôi nhé.
Chúc bạn học tốt!
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
a) \(5\frac{4}{23}.27\frac{3}{47}+4\frac{3}{47}\left(-5\frac{4}{23}\right)\)
\(=5\frac{4}{23}\left(27\frac{3}{47}-4\frac{3}{47}\right)\)
\(=5\frac{4}{23}\left(27+\frac{3}{47}-4-\frac{3}{47}\right)\)
\(=5\frac{4}{23}.23\)
\(=\frac{119}{23}.23=119\)
a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
= \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+\frac{1}{2}-\frac{36}{41}\)
= \(\frac{1}{2}-\left\{\frac{11}{24}+\frac{13}{24}\right\}-\left\{\frac{5}{41}+\frac{36}{41}\right\}\)
=\(\frac{1}{2}-\frac{24}{24}-\frac{41}{41}\)
=\(\frac{1}{2}-1-1\)
=\(\frac{-3}{2}\)
b) \(-12:\left\{\frac{3}{4}-\frac{5}{6}\right\}^2\)
= \(-12:\left\{\frac{9}{12}-\frac{10}{12}\right\}^2\)
= \(-12:\left\{\frac{-1}{12}\right\}^2\)
= \(-12:\frac{1}{144}\)
= \(-12.144\)
= -1728
c) \(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left[\left(\frac{-24}{18}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-23}{6}\right)\)
= \(\frac{-7}{6}\)
d) \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)
= \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)
= \(\left\{23\frac{1}{4}-13\frac{1}{4}\right\}.\frac{7}{5}\)
= \(10.\frac{7}{5}\)
= 14
e) (1+23−14).(0,8−34)2
= (1+23−14).(\(\frac{4}{5}\)−34)2
= \(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\)
= \(\frac{17}{12}.\left(\frac{1}{20}\right)^2\)
= \(\frac{17}{20}.\frac{1}{400}\)
= \(\frac{17}{8000}\)
a: ~12.58
b: 10
c:-1
d:55/4
dkr68