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1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
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) 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: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
Bài 1:
\(a,\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2-\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{24+2-3}{12}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\dfrac{23}{12}=\dfrac{7}{46}\)
\(x+\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{7}{46}.\dfrac{23}{12}\)
\(x+\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{7}{24}\)
\(x+\dfrac{1}{4}=\dfrac{7}{24}+\dfrac{1}{3}\)
\(x+\dfrac{1}{4}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}-\dfrac{1}{4}=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\)
\(b,\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{7}{10}\)
\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\)
\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{1}{6}\)
\(\dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\)
\(\dfrac{13}{21}+x=\dfrac{2}{7}\)
\(x=\dfrac{2}{7}-\dfrac{13}{21}=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
Bài 2:
\(a,\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)
\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(\dfrac{121}{12}-\dfrac{19}{2}\right)\)
\(=\dfrac{77}{18}:\dfrac{7}{12}\)
\(=\dfrac{22}{3}\)
\(b,1\dfrac{5}{18}-\dfrac{5}{18}.\left(\dfrac{1}{15}+1\dfrac{1}{12}\right)\)
\(=\dfrac{23}{18}-\dfrac{5}{18}.\dfrac{69}{60}\)
\(=\dfrac{23}{18}-\dfrac{23}{72}\)
\(=\dfrac{23}{24}\)
\(c,-\dfrac{1}{7}.\left(9\dfrac{1}{2}-8,75\right):\dfrac{2}{7}+0,625:1\dfrac{2}{3}\)
\(=\dfrac{-1}{7}.\dfrac{3}{4}:\dfrac{2}{7}+\dfrac{5}{8}:\dfrac{5}{3}\)
\(=-\dfrac{3}{8}+\dfrac{5}{8}:\dfrac{5}{3}\)
\(=-\dfrac{3}{8}+\dfrac{3}{8}\)
\(=\dfrac{0}{8}=0\)
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bn có thể giải cho mik mấy bài mà mik vừa đăng đc ko mik đang cần gấp