\(\frac{\left(-5\right)^{60}.30^5}{15^5.5^{61}}=?\)
Tính:
a,\(\frac{\left(-5\right)^{60}.30^5}{15^5.5^{61}}\)b,\(\frac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)
\(a)\frac{(-5)^{60}.30^5}{15^5.5^{61}}=\frac{(5.2.3)^5}{(5.3)^5.5}=\frac{5^5.2^5.3^5}{5^5.3^5.5} =\frac{2^5}{5}=\frac{32}{5}\)
\(b) \frac{(-3)^{10}.15^5}{25^3.(-9)^7}=\frac{(-3)^{10} .(3.5)^5}{(5^2)^3.[(-3).3]^7}=\frac{(-3)^{10}.3^5.5^5}{5^6.(-3)^7.3^7}=\frac{(-3)^3}{5.3^2}=\frac{-3}{5}\)
~ Hok tốt a~
Baif1:
a. \(\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)
c. \(\frac{6^3.2^5\left(-3\right)^{2
}}{\left(-2\right)^9.3^7}\)
Cứu với chiều nộp rồi
a. \(\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}=\frac{3^3.5^3.5^4}{\left(-3\right)^5.5^6}\)
\(=\frac{3^3.5^7}{\left(-3\right)^5.5^6}=\frac{5}{-9}\)
b. \(\frac{6^3.2^5.\left(-3\right)^2}{\left(-2\right)^9.3^7}=\frac{2^3.3^3.2^5.3^2}{\left(-2\right)^9.3^7}\)
\(=\frac{2^8.3^5}{\left(-2\right)^9.3^7}=\frac{1}{\left(-2\right).3^2}=-\frac{1}{18}\)
Bài 1: Thực hiện phép tính
S= \(\frac{3}{\left(1\cdot2\right)^2}+\frac{5}{\left(2\cdot3\right)^2}+...+\frac{61}{\left(30\cdot31\right)^2}\)
\(S=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+.......+\frac{61}{\left(30.31\right)^2}\)
\(=\frac{1}{1^2.2^2}+\frac{1}{2^2.3^2}+....+\frac{1}{30^2.31^2}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{30}-\frac{1}{31}\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-......-\left(\frac{1}{30}-\frac{1}{30}\right)-\frac{1}{31}\)
\(=1-\frac{1}{31}\\ =\frac{31}{31}-\frac{1}{31}=\frac{30}{31}\)
no mình nha
Tính giá trị các biểu thức sau:
a) \({\left( {\frac{3}{4}} \right)^{ - 2}}{.3^2}{.12^0}\);
b) \({\left( {\frac{1}{{12}}} \right)^{ - 1}}.{\left( {\frac{2}{3}} \right)^{ - 2}}\);
c) \({\left( {{2^{ - 2}}{{.5}^2}} \right)^{ - 2}}:\left( {{{5.5}^{ - 5}}} \right)\).
a) \(\left(\dfrac{3}{4}\right)^{-2}\cdot3^2\cdot12^0=16\)
b) \(\left(\dfrac{1}{12}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-2}=27\)
c) \(\left(2^{-2}\cdot5^2\right)^{-2}:\left(5\cdot5^{-5}\right)=16\)
\(\frac{\frac{2}{3}+3\left(\frac{2}{3}\right)^3-\left(\frac{5}{6}\right)^2}{\frac{7}{60}:\left(\frac{35}{31\cdot37}+\frac{35}{37\cdot43}+\frac{105}{43\cdot61}+\frac{35}{61\cdot67}\right)}\)
Giúp mình đi ai giúp đầu tặng 1 like
a) A=\(\left(\frac{-5}{11}\right).\frac{7}{15}.\left(\frac{11}{-5}\right).\left(-30\right)\)
b) B=\(\left(-\frac{1}{6}\right).\left(\frac{-15}{19}\right).\left(\frac{38}{45}\right)\)
c) C= \(\left(\frac{-5}{9}\right).\frac{3}{11}+\left(\frac{-13}{18}\right).\frac{3}{11}\)
d) D= \(\left(2\frac{2}{15}.\frac{9}{17}.\frac{3}{32}\right):\left(\frac{-3}{27}\right)\)
Thực hiện phép tính:
S = \(\frac{3}{\left(1\cdot2\right)^2}\)+\(\frac{5}{\left(2\cdot3\right)^2}\)+.......+\(\frac{61}{\left(30\cdot31\right)^2}\)
\(S=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{61}{\left(30.31\right)^2}\)
\(S=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{61}{30^2.31^2}\)
\(S=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{61}{900.961}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{900}-\frac{1}{961}\)
\(S=1-\frac{1}{961}\)
\(S=\frac{960}{961}\)
\(\frac{\left(3^2.5.7^9\right).\left(3^5.5^3\right)}{\left(3^3.5^2.7^5\right).^2}\)
= \(\frac{3^2.5^4.7^9}{3^3.5^2.7^5.3^3.5^2.7^5}\)
=\(\frac{3^2.5^4.7^9}{3^6.5^4.7^{10}}\)
= \(\frac{1.1.1}{3^4.1.7}\)
= \(\frac{1}{567}\)
\(\frac{\left(3^2.5.7^9\right).\left(3^5.5^3\right)}{\left(3^3.5^2.7^5\right)^2}=\frac{\left(3^2.3^5\right).\left(5.5^3\right).7^9}{3^6.5^4.7^{10}}=\frac{3^7.5^4.7^9}{3^6.5^4.7^{10}}=\frac{3}{7}\)
Tính giá trị biểu thức
\(1.A=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{3}\right)-\frac{1}{7}+\left[\frac{-14}{30}\right]\)
\(2.B=\left(\frac{5}{8}-\frac{4}{12}+\frac{3}{2}\right)-\left(\frac{5}{8}+\frac{9}{13}\right)-\left[\frac{-3}{2}\right]+\frac{7}{-15}\)
\(3.C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)
\(4.D=\frac{1}{9}-\left[\frac{-5}{23}\right]-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{14}-\frac{7}{30}\)
\(5.E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)+\left(\frac{12}{17}+\frac{5}{18}+\frac{7}{5}\right)\)
\(6.F=\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{12}{17}-\frac{15}{14}+\frac{1}{4}\right)\)
\(7.G=\frac{1}{25}-\frac{4}{27}+\left(\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}\right)+\frac{5}{43}-\frac{4}{7}\)
\(8.H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{29}{27}\right)-\frac{2}{27}\)
\(9.K=\frac{1}{16}-\frac{5}{21}+\left(\frac{-1}{16}+\frac{-3}{5}-\frac{-5}{21}\right)+\frac{-2}{5}+\frac{3}{4}\)
\(10.L=\frac{7}{12}+\frac{15}{14}-\left(\frac{14}{22}+\frac{-1}{14}+\frac{5}{21}\right)-\frac{-5}{21}+\frac{3}{5}\)
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