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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: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
Bài 1 : Rút gọn các phân số sau đến tối giản :
a) \(\dfrac{3.21}{14.15}=\dfrac{3.3.7}{2.7.3.5}=\dfrac{1.3.1}{2.1.1.5}=\dfrac{3}{10}\)
b) \(\dfrac{49+49.7}{49}=\dfrac{49\left(1+7\right)}{49}=\dfrac{49.8}{49}=\dfrac{1.8}{1}=\dfrac{8}{1}=8\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
Bài 1:
a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{2}{4}\)
\(=\dfrac{3}{4}\)
b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{5}{10}+\dfrac{8}{10}\)
\(=\dfrac{9}{5}\)
c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)
\(=\dfrac{1}{6}-\dfrac{7}{2}\)
\(=\dfrac{1}{6}-\dfrac{21}{6}\)
\(=\dfrac{-10}{3}\)
e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\dfrac{2}{3}\)
f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{3}{2}\)
\(=\dfrac{2}{2}=1\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{2}{4}-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{9}{28}\)
\(=\dfrac{196}{140}-\dfrac{45}{140}\)
\(=\dfrac{151}{140}\)
i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)
\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)
k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
a) \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
\(=\dfrac{1135}{23}-\left(\left(5+14\right)+\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\right)\)
\(=\dfrac{1135}{23}-\left(19+\dfrac{417}{736}\right)\)
\(=\dfrac{1135}{23}-19\dfrac{417}{736}\)
\(=\dfrac{1135}{23}-\dfrac{14401}{736}\)
\(=\dfrac{953}{32}\)
b) \(-\dfrac{3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=-\dfrac{1}{7}\cdot\dfrac{5}{3}-\dfrac{4}{3}\cdot\dfrac{1}{7}+\dfrac{17}{7}\)
\(=-\dfrac{5}{21}-\dfrac{4}{21}+\dfrac{17}{7}\)
\(=2\)
c) \(0,7\cdot2\dfrac{2}{3}\cdot20\cdot0,375\cdot\dfrac{5}{28}\)
\(=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)
\(=2\cdot\dfrac{5}{4}\)
\(=\dfrac{5}{2}\)
d) \(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(=\left(9\cdot\dfrac{3}{8}+\dfrac{303030}{69264}\right)+\dfrac{403}{100}\)
\(=\left(\dfrac{27}{8}+\dfrac{35}{8}\right)+\dfrac{403}{100}\)
\(=\dfrac{31}{4}+\dfrac{403}{100}\)
\(=\dfrac{589}{50}\)
P/s: Đánh dấu phẩy, dấu chấm (dấu nhân) cần rõ ràng (vì dấu chấm người ta sẽ hiểu là dấu nhân thay vì hiểu là dấu phẩy)
a) \(49\dfrac{8}{23}\)- \(\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
= \(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)+5\dfrac{7}{32}\)
=35+\(5\dfrac{7}{32}\)
=\(\dfrac{1287}{32}\)
b)\(-\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
=\(\left[\left(\dfrac{-3}{7}\right).\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\right]+2\dfrac{3}{7}\)
=\(\left[\left(\dfrac{-3}{7}\right).\dfrac{9}{9}\right]+2\dfrac{3}{7}\)
=\(\left[\left(\dfrac{-3}{7}\right).1\right]+2\dfrac{3}{7}\)
=\(\left(\dfrac{-3}{7}\right)+2\dfrac{3}{7}\)
=2
c) 0,7.\(2\dfrac{2}{3}\).20.0.375.\(\dfrac{5}{28}\)
=0 (Vì có một thừ số là 0 nên nguyên cả tích là 0)
d)\(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
=17+4,03
=21,03
a) \(\dfrac{1}{2}.\left(\dfrac{2}{9}+\dfrac{3}{7}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{41}{63}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\dfrac{88}{189}\)
\(=\dfrac{44}{189}\)
b) \(\left(\dfrac{-5}{28}+1,75+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\dfrac{-69}{20}\)
\(=\dfrac{-12}{23}\)
c) \(\dfrac{1}{3}.\dfrac{5}{7}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{2}{3}\)
\(=\dfrac{-3}{7}\)
d) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-\dfrac{352}{5}\)
\(=\dfrac{1}{10}\)
em không trả lời được câu hỏi của chị nhưng chị có thể giúp em đăng bài toán lên bằng cách nào không
a, \(3-2.\dfrac{1}{3}=3-\dfrac{2}{3}=\dfrac{7}{3}\)
b, \(-\dfrac{5}{7}.\dfrac{1}{35}+\left(-8\right):\dfrac{49}{6}\\ =-\dfrac{1}{49}+\left(-8\right).\dfrac{6}{49}=\dfrac{-1+\left(-8\right).6}{49}\\ =-\dfrac{49}{49}=-1\)
a) 3 - 2.\(\dfrac{1}{3}\)=3 - \(\dfrac{2}{3}\)=\(\dfrac{3}{1}\) - \(\dfrac{2}{3}\)=\(\dfrac{7}{3}\)
b) _\(\dfrac{5}{7}\). \(\dfrac{1}{35}\)+(-8) :\(\dfrac{49}{6}\)
=_ \(\dfrac{1}{49}\)+(-8) :\(\dfrac{49}{6}\)
=_\(\dfrac{1}{49}\)+ _ \(\dfrac{48}{49}\)
=-1