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3,
a) (−23+37):45+(−13+47):45
= \(-\frac{5}{21}:\frac{4}{5}+\frac{5}{21}:\frac{4}{5}\)
= \(\left(-\frac{5}{21}+\frac{5}{21}\right):\frac{4}{5}\)
= \(0:\frac{4}{5}=0\)
2,
a) \(\frac{-3}{4}\).\(\frac{12}{-5}\).(\(\frac{-25}{6}\))
= \(\frac{-3.4.3.\left(-5\right).5}{4.\left(-5\right).3.3}\)
= \(-5\)
b) (−2).\(\frac{-38}{21}\).\(\frac{-7}{4}\).(\(\frac{-3}{8}\))
= \(\frac{-2.\left(-38\right)\left(-7\right)\left(-3\right)}{\left(-7\right)\left(-3\right)\left(-2\right)\left(-2\right).8}\)
= \(\frac{19}{8}\)
c) (\(\frac{11}{12}:\frac{33}{16}\)).\(\frac{3}{5}\)
= \(\left(\frac{11}{12}.\frac{16}{33}\right).\frac{3}{5}\)
= \(\frac{4}{9}.\frac{3}{5}\)
= \(\frac{4}{15}\)
d) \(\frac{7}{23}\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-41}{10}\right)\)
= \(\frac{-287}{203}\)
3. Tính:
a) (\(\frac{-2}{3}+\frac{3}{7}\)):\(\frac{4}{5}\)+(\(\frac{-1}{3}+\frac{4}{7}\)):\(\frac{4}{5}\)
= (\(\frac{-2}{3}+\frac{3}{7}\)\(+\)\(\frac{-1}{3}+\frac{4}{7}\)) : \(\frac{4}{5}\)
= 0 : \(\frac{4}{5}\)
= 0
b) \(\frac{5}{9}\):(\(\frac{1}{11}-\frac{5}{22}\))+\(\frac{5}{9}\):(\(\frac{1}{15}-\frac{2}{3}\))
= \(\frac{5}{9}\): \(\frac{-3}{22}\)+ \(\frac{5}{9}\): \(\frac{-3}{5}\)
= \(\frac{5}{9}\): \(\frac{-81}{110}\)
= \(\frac{-550}{729}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
Câu a) số lớn lắm
b) \(3^{-3}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^x=3^3\)
=> \(3^x=3^3:\frac{1}{27}=3^3:\left(\frac{1}{3}\right)^3=3^3:\frac{1^3}{3^3}=3^3\cdot3^3=3^6\)
=> x = 6
b) \(\left(7x+2\right)^{-1}=3^{-2}\)
=> \(\frac{1}{7x+2}=\frac{1}{9}\)
=> 7x + 2 = 9
=> 7x = 7
=> x = 1
Bài 2:
a) \(3^4\cdot\frac{1}{729}\cdot81^3\cdot\frac{1}{9^2}\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot\left(3^4\right)^3\cdot\left(\frac{1}{3}\right)^4\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot3^{12}\cdot\left(\frac{1}{3}\right)^4=3^{16}\cdot\left(\frac{1}{3}\right)^{10}=\frac{3^{16}}{3^{10}}=3^6\)
b) \(\left(8\cdot2^5\right):\left(2^4\cdot\frac{1}{32}\right)=\left(2^3\cdot2^5\right):\left(2^4\cdot\left(\frac{1}{2}\right)^5\right)\)
\(=2^8:\left(2^4\cdot\frac{1^5}{2^5}\right)=2^8:\left(\frac{2^4}{2^5}\right)=2^8:2^{-1}=512\)
c) \(12^8\cdot9^{12}=\left(2^2\cdot3\right)^8\cdot\left(3^2\right)^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
d) Tương tự
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( 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) =
Tính:
a) \(-3\frac{3}{4}+\left(-\frac{10}{25}\right)+\left(-\frac{6}{12}\right)\)
\(=\left(-\frac{15}{4}\right)+\left(-\frac{10}{25}\right)+\left(-\frac{6}{12}\right)\)
\(=\left(-\frac{83}{20}\right)+\left(-\frac{6}{12}\right)\)
\(=-\frac{93}{20}.\)
b) \(-0,6-\left(-\frac{4}{9}\right)-\frac{16}{15}\)
\(=\left(-\frac{3}{5}\right)-\left(-\frac{4}{9}\right)-\frac{16}{15}\)
\(=\left(-\frac{7}{45}\right)-\frac{16}{15}\)
\(=-\frac{11}{9}.\)
Chúc bạn học tốt!
a) \(VT=12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
\(VP=18^{16}=2^{16}\cdot3^{32}\)
=> VT=VP
b) \(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^5}\)
(đề sai)
c) \(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)
\(VT=\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{3^6}{\left[3^3\left(3-1\right)\right]^2}=\frac{1}{2^2}=\frac{1}{4}=VP\)
128.912=186
=216.38.324=216.332
=216.332=186