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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) \(\frac{49}{5}\)
a, \(139\frac{5}{7}:\frac{2}{3}−138\frac{2}{7}:\sqrt{\frac{4}{9}} \)
= \(139\frac{5}{7}:\frac{2}{3}−138\frac{2}{7}:\frac{2}{3}\)
= \((139\frac{5}{7}−138\frac{2}{7}):\frac{2}{3}\)
= \(1\frac{3}{7}:\frac{2}{3}\)
= \(2\frac{1}{7}\)
b, \((\frac{-5}{11}:\frac{13}{18}-\frac{5}{11}:\frac{13}{5})+\frac{-1}{33} \)
= \((\frac{5}{11}.\frac{-18}{13}-\frac{5}{11}.\frac{5}{13})+\frac{-1}{33}\)
= \([\frac{5}{11}.(\frac{-18}{13}-\frac{5}{13})]+\frac{-1}{33}\)
= \((\frac{5}{11}.\frac{-23}{13})+\frac{-1}{33}\)
= \(\frac{-155}{143}+\frac{-1}{33}\)
= \(\frac{-358}{429} \)
c, \(∣97\frac{2}{3}-125\frac{3}{5}∣+97\frac{2}{3}-125\frac{3}{5} \)
= \(∣\frac{-419}{15}∣+97\frac{2}{3}-125\frac{3}{5}\)
= \(\frac{419}{15}+97\frac{2}{3}-125\frac{3}{5}\)
= \(0\)
Tick cho mình nha!!!
Chúc bạn học tốt.
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) \(\frac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^5}\)
\(=\frac{2\cdot\left(2\cdot3\right)^9-2^5\cdot\left(2\cdot3^2\right)}{2^2\cdot\left(2\cdot3\right)^5}\)
\(=\frac{2\cdot2^9\cdot3^9-2^5\cdot2^4\cdot3^8}{2^2\cdot2^5\cdot3^5}\)
\(=\frac{2^{10}\cdot3^9-2^9\cdot3^8}{2^7\cdot3^5}\)
\(=\frac{2^9\cdot3^8\left(2\cdot3-1\cdot1\right)}{2^7\cdot3^5}\)
\(=2^2\cdot3^3\left(2\cdot3-1\cdot1\right)\)
\(=4\cdot27\cdot5\)
\(=540\)
Câu c:
\(\sqrt{64}\) + 2\(\sqrt{\left(9-3\right)^2}\) - 7\(\sqrt{1,69}\) + 3\(\sqrt{\frac{25}{16}}\)
= 8 + 2.(9-3) - 7.1,3 + 3.\(\frac54\)
= 8 + 2.6 - 9,1 + 3.1,25
= 8 + 12 - 9,1 + 3,75
= 20 - 9,1 + 3,75
= 10,9 + 3,75
= 14,65
139\(\frac{5}{7}:\frac{2}{3}\)-\(138\frac{2}{7}:\sqrt{\frac{4}{9}}\)
=139\(\frac{5}{7}:\frac{2}{3}\)-\(138\frac{2}{7}:\frac{2}{3}\)
=(139\(\frac{5}{7}\)-\(138\frac{2}{7}\)):\(\frac{2}{3}\) =\(1\frac{3}{7}\):\(\frac{2}{3}\) =\(\frac{9}{7}.\frac{3}{2}\) =\(\frac{27}{14}\)=\(\frac{2.2^9.3^9-2^5.2^4.3^8}{2.2^8.3^8}\)
=\(\frac{2^{10}.3^9-2^9.3^8}{2^9.3^8}\)
=\(\frac{2^9.3^8.\left(2.3-1\right)}{2^9.3^8}\)
=\(6-1\)
=5