1. a) \(\dfrac{5}{3}+\dfrac{-2}{7}-\left(-1,2\right)\)
b) \(0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
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Những câu hỏi liên quan
31 tháng 5 2021
a) \(A=\dfrac{7}{8}\left(-\dfrac{5}{9}-\dfrac{4}{9}\right)+5\dfrac{7}{8}\)
\(A=\dfrac{7}{8}.\left(-1\right)+5\dfrac{7}{8}=5\dfrac{7}{8}-\dfrac{7}{8}=5\).
31 tháng 5 2021
\(B=\dfrac{1}{4}.\dfrac{8}{5}.\dfrac{25}{16}.\dfrac{-7}{4}=\dfrac{-35}{32}\)
TT
16 tháng 9 2017
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
PN
28 tháng 4 2017
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) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
PN
29 tháng 4 2017
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
25 tháng 12 2021
\(\text{A=}\left(\dfrac{7}{8}-0,25\right):\left(\dfrac{5}{6}-0,75\right)^2\)
\(A=\left(\dfrac{7}{8}-\dfrac{1}{4}\right):\left(\dfrac{5}{6}-\dfrac{3}{4}\right)^2\)
\(A=\left(\dfrac{7}{8}+\dfrac{-1}{4}\right):\dfrac{1}{144}\)
\(A=\dfrac{5}{8}.144=90\)
\(\text{B=}\dfrac{3}{4}.1\dfrac{4}{9}-\left(\dfrac{-3}{4}\right)\)
\(B=\dfrac{3}{4}.\dfrac{13}{9}+\dfrac{3}{4}\)
\(B=\dfrac{13}{12}+\dfrac{3}{4}\)
\(B=\dfrac{11}{6}\)
20 tháng 12 2021
e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
GV
Giáo viên Toán
Giáo viên
26 tháng 4 2017
a) \(\left(\dfrac{1}{16}\right)^{-\dfrac{3}{4}}+810000^{0.25}-\left(7\dfrac{19}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{4.\left(-\dfrac{3}{4}\right)}+\left(30\right)^{4.0,25}-\left(\dfrac{243}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{-3}+30-\left(\dfrac{3}{2}\right)^{5.\dfrac{1}{5}}\)
\(=2^3+30-\dfrac{3}{2}\)
\(=36,5\)
GV
Giáo viên Toán
Giáo viên
26 tháng 4 2017
b) \(=\left(0,1\right)^{3.\left(-\dfrac{1}{3}\right)}-2^{-2}.2^{6.\dfrac{2}{3}}-\left[\left(2\right)^3\right]^{-\dfrac{4}{3}}\)
\(=0,1^{-1}-2^2-2^{-4}\)
\(=10-4-\dfrac{1}{16}\)
\(=\dfrac{95}{16}\)
5 tháng 8 2021
a, \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)
\(=\dfrac{5}{7}+\dfrac{-5}{7}+\dfrac{3}{5}
=0+\dfrac{3}{5}=\dfrac{3}{5}\)
b, \(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2=\dfrac{-3}{4}-\dfrac{-3}{2}+1=\dfrac{-3}{4}-\dfrac{-6}{4}+1=\dfrac{3}{4}+1=\dfrac{7}{4}\)
c, \(\dfrac{-5}{9}+\left(\dfrac{-2}{3}\right)^2.\left(20\%-1.2\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(\dfrac{1}{5}-\dfrac{6}{5}\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(-1\right)=\dfrac{-5}{9}+\dfrac{-4}{9}=-1\)
VV
5 tháng 8 2021
Bài 1:
a) \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)\(=\left(\dfrac{5}{7}+\dfrac{-5}{7}\right)+\dfrac{3}{5}\)\(=0+\dfrac{3}{5}=\dfrac{3}{5}\)
b) \(\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2\)\(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+1\)\(=\dfrac{-3}{4}-\dfrac{-3}{2}+1\)
\(=\dfrac{3}{4}+1\)\(=\dfrac{7}{4}\)
DD
13 tháng 2 2022
A -\(\dfrac{24}{25}\)
B -\(\dfrac{5}{21}\)
C -\(\dfrac{24}{47}\)
D -\(\dfrac{19}{42}\)
tick cho mk
\(a,=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{271}{105}\\ b,=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-\dfrac{5}{4}=1-1-\dfrac{1}{8}=-\dfrac{1}{8}\)
1a) \(\dfrac{5}{3}-\dfrac{2}{7}+1,2=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}=\dfrac{107-30+126}{105}=\dfrac{203}{205}\)