\(\left(\dfrac{7}{30}\cdot\dfrac{12}{37}+\dfrac{12}{30}\cdot\dfrac{23}{37}\right)\cdot\dfrac{13}{31}-\dfrac{-25}{37}\cdot\dfrac{18}{31}\)
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TH
14 tháng 6 2018
Ta có:B=1\(\dfrac{6}{41}\)( \(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\) )
B=\(\dfrac{47}{41}\) [\(\dfrac{12\left(1+\dfrac{1}{19}-\dfrac{1}{37}-\dfrac{1}{53}\right)}{3\left(1+\dfrac{1}{3}-\dfrac{1}{37}-\dfrac{1}{53}\right)}:\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}\) B = \(\dfrac{47}{41}\) [ \(\dfrac{12}{3}:\dfrac{4}{5}\)]
B = \(\dfrac{47}{41}\)[ 4 . \(\dfrac{5}{4}\)]
B = \(\dfrac{47}{41}.5\)
B = \(\dfrac{235}{41}\)
Chúc bn hc tốt!!!
AH
Akai Haruma
Giáo viên
15 tháng 11 2017
Lời giải:
Ta có:
\(\text{VT}=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=\frac{1.2.3....31}{2.4.6.8...64}\)
Xét mẫu số:
\(2.4.6.8.....62.64=(2.1)(2.2)(2.3)(2.4)....(2.31)(2.32)\)
\(=2^{32}(1.2.3....31.32)\)
Suy ra:
\(\text{VT}=\frac{1.2.3....31}{2^{32}.(1.2.3...31.32)}=\frac{1}{2^{32}.32}=\frac{1}{2^{37}}\)
Do đó \(4^x=\frac{1}{2^{37}}\Leftrightarrow 2^{2x}=\frac{1}{2^{37}}\Leftrightarrow 2^{2x+37}=1\)
\(\Leftrightarrow 2x+37=0\Leftrightarrow x=-\frac{37}{2}\)
Vậy \(x=\frac{-37}{2}\)
5 tháng 4 2017
\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)
6 tháng 4 2017
\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)
17 tháng 5 2022
\(=\dfrac{7}{8}\left(\dfrac{2}{9}-\dfrac{1}{18}+\dfrac{1}{36}-\dfrac{5}{12}\right)=\dfrac{7}{8}\cdot\dfrac{8-2+1-15}{36}\)
\(=\dfrac{7}{8}\cdot\dfrac{-8}{36}=\dfrac{-7}{36}\)
30 tháng 12 2020
[3/7.4/15+1/3.(9^15)]^0.1/3.6^8/12^4
= 1.1/3.(2.3)^8/(3.4)^4
= 1/3.2^8.3^8/3^4.4^4
= 1/3.2^8.3^8/3^4.2^8
= 1/3.3^8/3^4
= 1/3.3^4=27
(dấu . là nhân nha)
NA
Nguyễn An Ninh
VIP
22 tháng 4 2023
=\(\dfrac{11}{31}\)*\(\left(\dfrac{-2}{17}-\dfrac{-9}{17}\right)\)+\(\dfrac{7}{31}\)*\(\dfrac{-20}{17}\)
=\(\dfrac{11}{31}\)*\(\dfrac{-7}{17}\)+\(\dfrac{7}{31}\)*\(\dfrac{-20}{17}\)
=\(\dfrac{-77}{512}\)+\(\dfrac{-140}{512}\)
=\(\dfrac{-217}{512}\)
\(\left(\dfrac{7}{30}\cdot\dfrac{12}{37}+\dfrac{12}{30}\cdot\dfrac{23}{37}\right)\cdot\dfrac{13}{31}-\dfrac{-25}{37}\cdot\dfrac{18}{31}\)
\(=\left(\dfrac{7}{30}\cdot\dfrac{12}{37}+\dfrac{12}{37}\cdot\dfrac{23}{30}\right)\cdot\dfrac{13}{31}+\dfrac{25}{37}\cdot\dfrac{18}{31}\)
\(=\dfrac{12}{37}\cdot\left(\dfrac{7}{30}+\dfrac{23}{30}\right)\cdot\dfrac{13}{31}+\dfrac{25}{37}\cdot\dfrac{18}{31}\)
\(=\dfrac{12}{37}\cdot1\cdot\dfrac{13}{31}+\dfrac{18}{37}\cdot\dfrac{25}{31}\)
\(=\dfrac{6}{37}\cdot\dfrac{26}{31}+\dfrac{6}{37}\cdot\dfrac{75}{31}\)
\(=\dfrac{6}{37}\cdot\left(\dfrac{26}{31}+\dfrac{75}{31}\right)\)
\(=\dfrac{6}{37}\cdot\dfrac{101}{31}\)
\(=\dfrac{606}{1147}\)