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Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
a: \(\frac{4^{30}\cdot3^{43}}{2^{53}\cdot27^{15}}\)
\(=\frac{\left(2^2\right)^{30}\cdot3^{43}}{2^{53}\cdot\left(3^3\right)^{15}}\)
\(=\frac{2^{60}}{2^{53}}\cdot\frac{3^{43}}{3^{45}}=\frac{2^7}{3^2}=\frac{128}{9}\)
b: \(\left(-27\right)^{10}\cdot16^{25}\)
\(=-3^{30}\cdot\left(2^4\right)^{25}=-3^{30}\cdot2^{100}\)
\(6^{30}\cdot\left(-32\right)^{15}\)
\(=2^{30}\cdot3^{30}\cdot\left(-1\right)\cdot\left(2^5\right)^{15}=-2^{105}\cdot3^{30}\)
\(\frac{\left(-27\right)^{10}\cdot16^{25}}{6^{30}\cdot\left(-32\right)^{15}}\)
\(=\frac{-2^{100}\cdot3^{30}}{-2^{105}\cdot3^{30}}=\frac{1}{2^5}=\frac{1}{32}\)
c: \(\left(-\frac{2}{17}\right)^3\cdot\left(-\frac{17}{35}\right)^3\cdot\left(\frac{35}{3}\right)^3\)
\(=\left(-\frac{2}{17}\cdot\frac{-17}{35}\cdot\frac{35}{3}\right)^3\)
\(=\left(\frac23\right)^3=\frac{8}{27}\)
e: \(2^2+4^2+\cdots+24^2\)
\(=2^2\left(1^2+2^2+\cdots+12^2\right)\)
\(=4\cdot650=2600\)
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!
\(M=\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{17}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)
\(M=\frac{\left(\frac{1}{30}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{24}{119}+\frac{3}{35}\right).\frac{-4}{3}}\)
\(M=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{171}{595}.\frac{-4}{3}}\)
\(M=\frac{-1}{12}:\frac{-228}{595}\)
\(M=\frac{595}{2736}\)
Ta có:
\(M=\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right)\times\frac{5}{19}}{\left(\frac{1}{17}+\frac{1}{7}-\frac{-3}{35}\right)\times\frac{-4}{3}}\)
\(M=\frac{\left(\frac{1}{30}-\frac{7}{20}\right)\times\frac{5}{19}}{\left(\frac{24}{119}+\frac{3}{35}\right)\times\frac{-4}{3}}\)
\(M=\frac{\frac{-19}{60}\times\frac{5}{19}}{\frac{171}{595}\times\frac{-4}{3}}\)
\(M=\frac{-1}{12}\div\frac{-228}{595}\)
\(M=\frac{595}{2736}\)
Vậy \(M=\frac{595}{2736}\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)