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a) Hình như nhầm đề thì phải :v
\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{6}{11}}\)
\(=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{5}{11}}=1\)
b) \(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(0,125-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}\right)}=\dfrac{1}{3}+\dfrac{1}{\dfrac{3}{2}}\)
\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)
a,\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}=\dfrac{\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right).132}{\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right).132}=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
b, Ta có : 0,125 = \(\dfrac{1}{8}\) ; 0,375 = \(\dfrac{3}{8}\) ; 0,2 = \(\dfrac{1}{5}\) ; 0,5 = \(\dfrac{3}{6}\)
\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\cdot\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{2\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}{3\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}\)
\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)
Ta có : \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
\(=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}\right).\frac{2005}{1890}+115\)
\(=\left(\frac{3}{5}-\frac{3}{5}\right).\frac{2005}{1890}+115=0+115=115\)
= ( \(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)+ \(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)) x \(\frac{2005}{1890}\)+ 115
= ( \(\frac{3(\frac{1}{2}+\frac{1}{3}-\frac{1}{4})}{5(\frac{1}{1}+\frac{1}{3}-\frac{1}{4})}\)+ \(\frac{3(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}{-5(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}\)) x \(\frac{2005}{1890}\)+ 115
=( \(\frac{3}{5}\)+\(\frac{3}{-5}\)) x \(\frac{2005}{1890}\)+115 = 0 +115 = 115
A=−85+105−115−12583−103+113+123+25+35−4523+33−43A=−5(81−101+111+121)3(81−101+111+121)+5(21+31−41)3(21+31−41)A=5−3+53=0
A=−85+105−115−12583−103+113+123+25+35−4523+33−43A=−5(81−101+111+121)3(81−101+111+121)+5(21+31−41)3(21+31−41)A=5−3+53=0
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\\ =\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\\ =\dfrac{1}{2}+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\dfrac{4}{5}\\ =\dfrac{1}{2}+0+\dfrac{4}{5}\\ =\dfrac{1}{2}+\dfrac{4}{5}\\ =\dfrac{5}{10}+\dfrac{8}{10}\\ =\dfrac{13}{10}\)
\(\dfrac{-3}{7}+\dfrac{3}{4}:\dfrac{3}{14}\\ =\dfrac{-3}{7}+\dfrac{3}{4}\cdot\dfrac{14}{3}\\ =\dfrac{-3}{7}+\dfrac{7}{2}\\ =\dfrac{-6}{14}+\dfrac{49}{14}\\ =\dfrac{43}{14}\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
a) \(\left(6\dfrac{1}{9}+3\dfrac{7}{11}\right)-1\dfrac{1}{9}\)
\(=\left(\left(6+3\right)+\left(\dfrac{1}{9}+\dfrac{7}{11}\right)\right)-\dfrac{10}{9}\)
\(=\left(9+\dfrac{74}{99}\right)-\dfrac{10}{9}\)
\(=9\dfrac{74}{99}-\dfrac{10}{9}\)
\(=\dfrac{965}{99}-\dfrac{10}{9}\)
\(=\dfrac{95}{11}\)
b) \(1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:5\%\)
\(=1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:\dfrac{1}{20}\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+3\cdot20\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+60\)
\(=\dfrac{1275}{22}\)
c) \(4\dfrac{3}{4}-0,37+\dfrac{1}{8}-1,28-2,5+3\cdot\dfrac{1}{2}\)
\(=\dfrac{19}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{3}{2}\)
\(=\dfrac{89}{40}\)
d) \(\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-5\cdot\dfrac{2}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-\dfrac{10}{11}\)
\(=-\dfrac{25}{77}\)
e) \(\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
(câu này chờ mình một chút)
Câu e) anh rút thừa số chung là 2 cùng mũ số ra ngoài thì phân số thành số nguyên. Em nghĩ thế!
a) \(1\dfrac{7}{20}:2,7+2,7:1,35+\left(0,4:2\dfrac{1}{2}\right).\left(4,2-1\dfrac{3}{40}\right)\)
\(=\dfrac{27}{20}:\dfrac{27}{10}+\dfrac{27}{10}:\dfrac{27}{10}+\left(\dfrac{2}{5}:\dfrac{5}{2}\right).\left(\dfrac{21}{5}-\dfrac{43}{40}\right)\)
\(=1+1+\dfrac{4}{25}.\dfrac{25}{8}\)
\(=2+\dfrac{1}{2}\)
\(=2\dfrac{1}{2}\)
b) \(\left(6\dfrac{3}{5}-3\dfrac{3}{14}\right).5\dfrac{5}{6}:\left(21-1.25\right):2,5\)
\(=\left(\dfrac{33}{5}-\dfrac{45}{14}\right).\dfrac{35}{6}:\left(-4\right):2,5\)
\(=\left(\dfrac{462}{60}-\dfrac{225}{60}\right).\dfrac{35}{6}.\dfrac{1}{-4}:\dfrac{5}{2}\)
\(=\dfrac{237}{60}.\dfrac{35}{6}.\dfrac{1}{-4}.\dfrac{2}{5}\)
\(=\dfrac{3.79.7.5.2}{5.14.3.2.\left(-4\right).5}\)
\(=\dfrac{79.7}{14.\left(-4\right).5}=\dfrac{553}{-280}\) (số xấu :v)
a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)
\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)
\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)
d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ A=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ A=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}=\dfrac{3\left(0,125-0,1+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(0,125-0,1+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{5}\left(2,5+\dfrac{5}{3}-1,25\right)}{2,5+\dfrac{5}{3}-1,25}=-\dfrac{3}{5}+\dfrac{3}{5}=0\)