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a) -( -35 + 7 - 13 ) + ( 2 - 35 ) - ( 13 - 7 )
= 35 - 7 + 13 + 2 - 35 - 13 + 7
= ( 35 - 35 ) + ( -7 + 7 ) + ( 13 - 13 ) + 2
= 0 + 0 + 0 + 2
= 2
b) ( 70 + 8 - 35 ) - ( 8 - 37 - 10 ) - 70
= 70 + 8 - 35 - 8 + 37 + 10 - 70
= ( 70 - 70 ) + ( 8 - 8 ) + ( -35 + 37 ) + 10
= 0 + 0 + 2 + 10
= 12
c) 5 - ( 14 + 4 - 52 ) - ( 52 - 14 ) + 4
= 5 - 14 - 4 + 52 - 52 + 14 + 4
= 5 - ( 14 + 14 ) + ( 52 - 52 ) + ( -14 + 14 )
= 5 - 14 - 14 + 0 + 0
= 5 - ( 14 - 14 )
= 5 - 0
= 5
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
a) 38 : 34 + 22 x 23 = 113
b) 3 x 42 - 2 . 3 = 42
c) 46 . 34 . 95 : 612 = 9
d) 212 . 14 . 125 : 353 . 6 = 108
e) 453 . 204 .182 : 1805 = 25
#BạcHà#
bn ơi ghi cả cách làm nữa bn
ai lm lun cách làm mk sẽ k cho 2 k lun
Giải: (^ là mũ)
a) 3^8 : 3^4 + 2^2 . 2^3
= 6561 : 81 + 4 . 8
= 81 + 32
= 113
Mấy câu sau bn làm tương tự nhé!
b: \(=2^{12}\cdot3^4\cdot\dfrac{3^{10}}{3^{12}\cdot2^{12}}=3^2=9\)
c: \(=\dfrac{7^2\cdot3^2\cdot7\cdot2\cdot5^3}{5^3\cdot7^3\cdot2\cdot3}=\dfrac{3^2}{3}=3\)
d: \(=\dfrac{5^3\cdot3^6\cdot5^4\cdot2^8\cdot3^{16}\cdot2^2}{\left(2^2\cdot3^2\cdot5\right)^5}=\dfrac{5^{10}\cdot3^{22}\cdot2^{10}}{2^{10}\cdot3^{10}\cdot5^5}=3^{12}\cdot5^5\)
\(a,\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+\frac{17}{-35}\)
=\(-\frac{5}{13}+\left(\frac{3}{17}+\frac{14}{17}\right)+\left(\frac{-18}{35}+\frac{-17}{35}\right)\)
= \(-\frac{5}{13}+1+\left(-1\right)\)
=\(-\frac{5}{13}\)
\(b,\frac{-3}{8}.\frac{1}{6}+\frac{3}{-8}.\frac{5}{6}+\frac{-10}{6}\)
=\(\frac{-3}{8}.\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{-10}{6}\)
=\(\frac{-3}{8}.1+\frac{-10}{6}\)
=\(-\frac{49}{24}\)
\(c,\frac{-4}{11}.\frac{5}{15}.\frac{11}{-4}\)
=\(\left(\frac{-4}{11}.\frac{11}{-4}\right).\frac{1}{3}\)
=\(1.\frac{1}{3}=\frac{1}{3}\)
\(d,\frac{13}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}\)
=\(\frac{13}{8}+\frac{1}{2}+\frac{-1}{8}\)
=\(\left(\frac{13}{8}+\frac{-1}{8}\right)+\frac{1}{2}\)
=\(\frac{3}{2}+\frac{1}{2}=2\)
\(e,\frac{-1}{2^2}-\left(-2\right)^2-5\)
=\(\frac{-1}{4}-4-5\)
=\(-\frac{37}{4}\)
\(f,\frac{121}{3}-\frac{5}{7}:\left(24-\frac{23}{57}\right)\)
=\(\frac{121}{3}-\frac{5}{7}:\frac{1345}{57}\)
=\(\frac{121}{3}-\frac{57}{1883}\)
\(\approx40,4\)