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a, A = \(\dfrac{3^{10}\times10+3^{10}\times6}{3^9\times2^4}\)
A = \(\dfrac{3^{10}\times\left(10+6\right)}{3^9\times2^4}\)
A = \(\dfrac{3^{10}\times16}{3^9\times16}\)
A = 3
c, C = \(\dfrac{36^{10}\times25^{15}}{30^8}\)
C = \(\dfrac{\left(6^2\right)^{10}.\left(5^2\right)^{15}}{30^8}\)
C = \(\dfrac{6^{20}.5^{30}}{6^8.5^8}\)
C = 612.522
a) 2x + (-61) - (21 - 61) = 2x - 21 + (61 - 61) = 2x - 21.
b) (- 3 - x + 5) + 3 = (- 3 + 3) + 5 - x = 5 - x.
c) 11- (13 - x) + (13 - 11) = (11- 11) + (13- 13) + x = x
d) 25 - ( 15 - x + 303) + 303 = 25 - 15 + (303 - 303) + x = x + 10
a, 15 . { 32 : [ 6 - 5 + 5 ( 9 : 3 ) ] + 3 } - 2018 0
= 15.{32:[1+15]+3}–1
= 15.5–1
= 74
b, 25 . { 2 7 : [ 12 - 4 + 2 2 . 16 : 2 3 ] - 2 4 }
= 25.{128:[8+4.2]–16}
= 25.24
= 600
c, 2019 . { 101 - 1000 : [ 2 2 . 2 3 + 5 6 : 5 3 - 6 2 : 11 - 2018 0 ] }
= 2019.{101–1000:[(32+125–36):11–1]}
= 2019.{101–1000:[121:11–1]}
= 2019.{101–1000:10}
= 2019.1
= 2019
\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)
\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)
Dễ thế cũng hỏi ở trường tính hay lắm mà:
a) \(\dfrac{3}{7}.\dfrac{8}{11}+\dfrac{3}{7}.\dfrac{5}{11}-\dfrac{3}{7}.\dfrac{2}{11}\)
\(=\dfrac{3}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=\dfrac{3}{7}.\left(\dfrac{8+5-2}{11}\right)=\dfrac{3}{7}.\dfrac{11}{11}=\dfrac{3}{7}\)
b) \(\dfrac{3}{13}.\dfrac{8}{25}+\dfrac{3}{13}.\dfrac{11}{25}+\dfrac{3}{13}.\dfrac{6}{25}-\dfrac{3}{13}\)
\(=\dfrac{3}{13}.\left(\dfrac{8}{25}+\dfrac{11}{25}+\dfrac{6}{25}-1\right)\)
\(=\dfrac{3}{13}.\left(\dfrac{8+11+6}{25}-1=\right)\dfrac{3}{13}.0=0\)
a,
\(\dfrac{3}{7}.\dfrac{8}{11}+\dfrac{3}{7}.\dfrac{5}{11}-\dfrac{3}{7}.\dfrac{2}{11}=\dfrac{3}{7}\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)=\dfrac{3}{7}.1=\dfrac{3}{7}\)
b,
\(\dfrac{3}{13}.\dfrac{8}{25}+\dfrac{3}{13}.\dfrac{11}{25}+\dfrac{3}{13}.\dfrac{6}{25}-\dfrac{3}{13}=\dfrac{3}{13}\left(\dfrac{8}{25}+\dfrac{11}{25}+\dfrac{6}{25}-1\right)=\dfrac{3}{13}.0=0\)