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Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
\(A=5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9\)
\(A=5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9\)
\(A=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}\)
\(A=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}\)
\(A=2^{29}\cdot3^{18}\cdot\left(5\cdot2^1\cdot1-1\cdot3^2\right)\)
\(A=2^{29}\cdot3^{18}\cdot\left(5-9\right)\)
\(A=-2^2\cdot2^{29}\cdot3^{18}\)
\(A=-2^{31}\cdot3^{18}\)
_______________
\(B=5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6\)
\(B=5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6\)
\(B=5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(B=2^{28}\cdot3^{18}\cdot\left(5\cdot1\cdot3-7\cdot2\cdot1\right)\)
\(B=2^{28}\cdot3^{18}\cdot\left(15-14\right)\)
\(B=2^{28}\cdot3^{18}\)
Ta có: \(A:B\)
\(=\left(-2^{31}\cdot3^{18}\right):\left(2^{28}\cdot3^{18}\right)\)
\(=\left(-2^{31}:2^{28}\right)\cdot\left(3^{18}:3^{18}\right)\)
\(=-2^3\cdot1\)
\(=-8\)
a) 15 . (27 + 18 + 6 ) + 15 . ( 23 +12 ) = 15. 51 + 15 . 35 = 765 + 525 = 1290
b) 14 . 35 . 5 + 10 . 25 . 7 + 20 . 70 = 2450 + 1750 + 1400 = 2452450
c) 53 . ( 51 + 4 ) + 53 . ( 49 + 96 ) + 53 = 53 . 55 + 53 . 145 + 53 = 2915 . 7685 = 22401775
d) 46 . ( 27 +75 ) +8 . ( 58 + 25 ) + 8 . ( 58 + 25 ) . 2 = 46 . 102 + 8 . 83 + 8 . 83 = 4692 + 664 + 664 = 6020
a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
a, 2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= 2 x 12 x 31 + 4 x 6 x 42 + 8 x 3 x 27
= 24 x 31 + 24 x 42 + 24 x 27
= 24 x ( 31 + 24 + 27 )
= 24 x 82
= 1968
b, 2 x 53 x 12 + 4 x 6 x 87 - 3 x 8 x 40
= 24 x 53 + 24 x 87 - 24 x 40
= 24 x ( 53 + 87 - 40 )
= 24 x 100
= 2400
c, Tương tự
\(a, 2023 - 25^2 : 5^3 + 27\)
\(=\left(2023+27\right)-\left(5^2\right)^2:5^3\)
\(=2050-5^4:5^3\)
\(=2050-5\)
\(=2045\)
___
\(b,60:\left[7\cdot\left(11^2-20\cdot6\right)+5\right]\)
\(=60:\left[7\cdot\left(121-120\right)+5\right]\)
\(=60:\left(7\cdot1+5\right)\)
\(=60:\left(7+5\right)\)
\(=60:12\)
\(=5\)
#Toru
a) \(2023-25^2:5^3+27\)
\(=2023-5^4:5^3+27\)
\(=2023-5+27\)
\(=2018+27\)
\(=2045\)
b) \(60:\left[7.\left(11^2-20.6\right)+5\right]\)
\(=60:\left[7.\left(121-120\right)+5\right]\)
\(=60:\left[7.1+5\right]\)
\(=60:\left[7+5\right]\)
\(=60:12\)
\(=5\)
\(#WendyDang\)