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\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow[\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)]:2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow1\dfrac{43}{45}:2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow\dfrac{88}{45}:\dfrac{8}{3}=0,01x:4\)
\(\Leftrightarrow\dfrac{88}{45}.\dfrac{3}{8}=0,01x:4\)
\(\Leftrightarrow\dfrac{11}{15}=0,01x:4\)
\(\Leftrightarrow0,01x:4=\dfrac{11}{15}\)
\(\Leftrightarrow x=\dfrac{11}{15}.4:0,01\)
\(\Leftrightarrow x=\dfrac{880}{3}\)
Vậy x = \(\dfrac{880}{3}\)
Ta có:
\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0.01x:4\\ \left(\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)\right):\dfrac{8}{3}=0.01x:4\\ \dfrac{88}{45}:\dfrac{8}{3}=0.01x:4\\ \dfrac{11}{15}=0.01x:4\\ \dfrac{44}{15}=0.01x\\ x=\dfrac{44}{15}:0.01\\ x=\dfrac{880}{3}\)
Vậy \(x=\dfrac{880}{3}\)
\(\dfrac{\text{45^{10^{ }}}.5^{10}}{75^{10}}=\dfrac{9^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}=\dfrac{9^{10}}{3^{10}}=3^{10}\)
\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)
Ta có : a, 25/7 + 13/21 - 11/7 + 17/21 + 1/3 .
= ( 25/7 - 11/7 ) + ( 13/21 + 17/21 + 1/3 ) .
= 2 + ( 20/21 + 7/21 ) .
= 2 + 9/7 .
= 23/7 .
b, ( 1/3 + 12/67 + 13/41 ) - ( 79/67 - 28/41 ) .
= 1/3 + 12/67 + 13/41 - 79/67 + 28/41 .
= 1/3 + ( 12/67 - 79/67 ) + ( 13/41 + 28/41 ) .
= 1/3 - 1 + 1 .
= 1/3 .
c, ( 11/4 . -5/9 - 4/9 . 11/4 ) . 8/33 .
= [ 11/4 . ( -5/9 - 4/9 ) ] . 8/33 .
= [ 11/4 . ( - 1 ) ] . 8/33 .
= -11/4 . 8/33 .
= -2/3 .
d, 38/45 - ( 8/45 - 17/51 - 3/11 ) .
= 38/45 - 8/45 + 17/51 + 3/11 .
= 2/3 + 17/51 + 3/11 .
= 374/561 + 187/561 + 153/561 .
= 14/11 .
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
a: \(=5-2\cdot\dfrac{1}{4}=5-\dfrac{1}{2}=\dfrac{9}{2}\)
b: \(=\left(\dfrac{7}{2}\right)^3+\dfrac{1}{2}=\dfrac{343}{8}+\dfrac{1}{2}=\dfrac{347}{8}\)
c: \(=\left(5+\dfrac{5}{27}-\dfrac{5}{27}\right)+\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\dfrac{1}{2}=5+1-\dfrac{1}{2}=5+\dfrac{1}{2}=5.5\)
e: \(=\dfrac{-5}{4}\left(35+\dfrac{1}{6}-45-\dfrac{1}{6}\right)=\dfrac{-5}{4}\cdot\left(-10\right)=\dfrac{50}{4}=\dfrac{25}{2}\)
a: \(=\dfrac{3}{8}\left(72+\dfrac{1}{5}-51-\dfrac{1}{5}\right)=\dfrac{3}{8}\cdot21=\dfrac{63}{8}\)
b: \(=25\cdot\dfrac{-1}{125}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{2}-\dfrac{1}{2}=-1\)
c: \(=4\left(35+\dfrac{1}{6}\right)\cdot\dfrac{-1}{5}-\left(45+\dfrac{1}{6}\right)\cdot\dfrac{-1}{5}\)
\(=\dfrac{-1}{5}\left(140+\dfrac{2}{3}-45-\dfrac{1}{6}\right)=-\dfrac{191}{10}\)
a. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
b. \(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-1728\)
c. \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]\)
\(=\dfrac{7}{23}.\dfrac{-23}{6}\)
\(=\dfrac{-7}{6}\)
d. \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}.\dfrac{7}{5}\)
\(=\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right).\dfrac{7}{5}\)
\(=10.\dfrac{7}{5}\)
\(=14\)
e. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)
\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)
a, \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.2^8.5^4}{5^{10}.2^{10}}=\dfrac{1}{5^2.2^2}=\dfrac{1}{25.4}=\dfrac{1}{100}\)
b, \(\dfrac{2^7.9^3}{6^5.8^2}=\dfrac{2^7.3^6}{2^5.3^5.2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
c, \(\dfrac{45^{10}.5^{20}}{75^5}=\dfrac{5^{10}.3^{20}.5^{20}}{3^5.5^{10}}=5^{20}.3^{15}\)
d, \(\left(0,8\right)^5=\left(0,1\right)^5.8^5=\dfrac{1}{100000}.32768=0,32768\)
e, \(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{2^{15}.3^8}{2^6.3^6.2^9}=3^2=9\)
d, \(\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}.\left(2^{20}+1\right)}{2^{30}.\left(2^{20}+1\right)}=2^{10}=1024\)
Chúc bạn học tốt!!!
\(\text{a) }\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot\left(5\cdot4\right)^4}{\left(5^2\right)^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^{10}\cdot4^5}=\dfrac{5^8\cdot4^4}{5^{10}\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{25\cdot4}=\dfrac{1}{100}\)
\(\text{b) }\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{2^7\cdot3^6}{2^5\cdot2^6\cdot3^5}=\dfrac{2^7\cdot3^6}{2^{11}\cdot3^5}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
\(\text{c) }\dfrac{45^{10}\cdot5^{20}}{75^5}=\dfrac{\left(5\cdot9\right)^{10}\cdot5^{20}}{\left(25\cdot3\right)^5}=\dfrac{5^{10}\cdot9^{10}\cdot5^{20}}{25^5\cdot3^5}=\dfrac{5^{10}\cdot5^{20}\cdot\left(3^2\right)^{10}}{\left(5^2\right)^5\cdot3^5}=\dfrac{5^{30}\cdot3^{20}}{5^{10}\cdot3^5}=5^{20}\cdot3^{15}\)
\(\text{d) }\left(0.8\right)^5=\left(\dfrac{8}{10}\right)^5=\left(\dfrac{4}{5}\right)^5=\dfrac{4^5}{5^5}=\dfrac{64}{3125}\)
\(\text{e) }\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^6\cdot2^9\cdot3^6}=\dfrac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=3^2=9\)
\(f\text{) }\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)
\(\dfrac{-2}{45}:\dfrac{8}{3}=0,01x:4\)
\(0,01x:4=\dfrac{-2}{45}\times\dfrac{3}{8}\)
\(0,01x:4=\dfrac{-6}{360}\)
\(0,01x:4=\dfrac{-1}{60}\)
\(0,01x=\dfrac{-1}{60}\times\dfrac{1}{4}\)
\(0,01x=\dfrac{-1}{240}\)
\(x=\dfrac{-1}{240}:0,01\)
\(x=\dfrac{-1}{240}\times100\)
\(x=\dfrac{-100}{240}=-2,4\)
\(\dfrac{-2}{45}:\dfrac{8}{3}=0,01x:4\)
\(\dfrac{-2}{45}\times\dfrac{3}{8}=0,01x:4\)
\(\dfrac{-6}{360}=0,01x:4\)
\(\dfrac{-1}{60}=0,01x:4\)
\(0,01x=\dfrac{-1}{60}\times\dfrac{4}{1}\)
\(0,01x=\dfrac{-4}{60}\)
\(0,01x=\dfrac{-1}{15}\)
\(x=\dfrac{-1}{15}:0,01\)
\(x=\dfrac{-1}{15}\times100\)
\(x=\dfrac{-100}{15}=\dfrac{-20}{3}\)
Vậy \(x=\dfrac{-20}{3}\).