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a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)
= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)
b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)
\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)
\(=66+44+33+22+12=177\)
c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)
= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)
= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)
= còn lại làm nốt nha! bận ròy
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
a) \(\left(\frac{11}{12}:\frac{44}{16}\right).\left(\frac{-1}{3}+\frac{1}{2}\right)\) \(=\left(\frac{11}{12}.\frac{16}{44}\right).\left(\frac{-2}{6}+\frac{3}{6}\right)\) \(=\frac{1}{3}.\frac{1}{6}\) \(=\frac{1}{18}\)
b) \(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}\) \(=\frac{\left(-5\right)^5.2^4}{5^4.\left(-2\right)^4}\) \(=5\) (Có sửa đề lại, nếu có sai thì ib mình sửa lại nhé!)
c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\) \(=\frac{15}{2}.\left(\frac{-3}{5}\right)+\frac{5}{2}.\left(\frac{-3}{5}\right)\) \(=\frac{-3}{5}.\left(\frac{15}{2}+\frac{5}{2}\right)\)
\(=\frac{-3}{5}.10\) \(=-6\)
d) \(\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{5}{4}\) \(=\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right).\frac{4}{5}\)
\(=\frac{4}{5}.\left(\frac{-1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\) \(=\frac{4}{5}.\left(\frac{0}{2}+1\right)\) \(=\frac{4}{5}.1=\frac{4}{5}\)
a) (1112:4416).(−13+12)(1112:4416).(−13+12) =(1112.1644).(−26+36)=(1112.1644).(−26+36) =13.16=13.16 =118=118
b) (−5)2.(−5)3.1654.(−2)4(−5)2.(−5)3.1654.(−2)4 =(−5)5.2454.(−2)4=(−5)5.2454.(−2)4 =5=
c) 7,5:(−53)+212:(−53)7,5:(−53)+212:(−53) =152.(−35)+52.(−35)=152.(−35)+52.(−35) =−35.(152+52)=−35.(152+52)
=−35.10=−35.10 =−6=−6
d) (−12+13).45+(23+12):54(−12+13).45+(23+12):54 =(−12+13).45+(23+12).45=(−12+13).45+(23+12).45
=45.(−12+13+23+12)=45.(−12+13+23+12) =45.(02+1)=45.(02+1) =45.1=45
a) \(\left(\frac{11}{4}.\frac{-5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)
=\(\frac{11}{4}\left(-\frac{5}{9}-\frac{4}{9}\right).\frac{8}{33}\)
=\(\frac{11}{4}\cdot-1\cdot\frac{8}{33}\)
=\(-\frac{11}{4}\cdot\frac{8}{33}\)
=\(-\frac{2}{3}\)
b)\(-\frac{1}{4}\cdot\frac{152}{11}+\frac{68}{4}\cdot-\frac{1}{11}\)
=\(\frac{-1.152}{4.11}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1.152}{11.4}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1}{11}\cdot\frac{152}{4}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1}{11}\cdot\left(\frac{152}{4}+\frac{68}{4}\right)\)
=\(\frac{-1}{11}\cdot55=-5\)
c)\(\frac{-2}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{3}{5}\)
=\(-1\cdot\frac{2}{3}\left(\frac{4}{5}+\frac{3}{5}\right)\)
=\(-1\cdot\frac{2}{3}\cdot\frac{7}{5}\)
=\(-\frac{2}{3}\cdot\frac{7}{5}\)
=\(\frac{-14}{15}\)
d) chưa nghĩ ra nhé
e) bạn chép sai đề bài rồi
mk mới kiểm tra 45 phút nên biết
đề bài nè
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{899}{30^2}\)
=\(\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot\frac{3.5}{4^2}\cdot...\cdot\frac{29.31}{30^2}\)
=\(\frac{1.3.2.4.3.5...29.31}{2.2.3^2.4^2...30.30}\)
=\(\frac{1.2.3^2.4^2.5^2....29^2.30.31}{2.2.3^2.4^2.5^2....29^2.30.30}\)
=\(\frac{1.31}{2.30}\)
=\(\frac{31}{60}\)
a)trong ngoac bn dat thau so chung la 11/4 rui tinh binh thuong b)bn tu lam nhe c)dat thua so chung d)tinh trong ngoac ra rui nhan vs e) mk bo tay
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
a) \(\frac{2}{3}+\frac{1}{3}\cdot\left(-\frac{2}{5}\right)\\ =\frac{2}{3}+\frac{-2}{15}\\ =\frac{10}{15}+\frac{-2}{15}\\ =\frac{8}{15}\)
b) \(0,75\cdot1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\\ =\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}\cdot\frac{-20}{21}\\ =\frac{4}{3}-\frac{-4}{3}\\ =\frac{4}{3}+\frac{4}{3}\\ =\frac{4}{3}\cdot2\\ =\frac{8}{3}\)
c) \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}-\frac{-4}{19}+\frac{8}{23}\\ =\frac{-2}{17}+\frac{15}{23}+\frac{-15}{17}+\frac{4}{19}+\frac{8}{23}\\ =\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\\ =\left(-1\right)+1+\frac{4}{19}\\ =0+\frac{4}{19}\\ =\frac{4}{19}\)
d) \(2019^0\cdot\left(6-2\frac{4}{5}\right)\cdot3\frac{1}{8}-1\frac{3}{5}:25\%\\ =1\cdot\left(\frac{30}{5}-\frac{14}{5}\right)\cdot\frac{25}{8}-\frac{8}{5}:\frac{1}{4}\\ =1\cdot\frac{16}{5}\cdot\frac{25}{8}-\frac{8}{5}\cdot4\\ =\frac{16}{5}\cdot\frac{25}{8}-\frac{32}{5}\\ =\frac{50}{5}-\frac{32}{5}\\ =\frac{18}{5}\)
e) \(\left(\frac{7}{8}-\frac{1}{2}\right)\cdot2\frac{2}{3}-\frac{3}{7}\cdot\left(2,5^2\right)\\ =\left(\frac{7}{8}-\frac{4}{8}\right)\cdot\frac{8}{3}-\frac{3}{7}\cdot6,25\\ =\frac{3}{8}\cdot\frac{8}{3}-\frac{3}{7}\cdot\frac{25}{4}\\ =1-\frac{75}{28}\\ =\frac{28}{28}-\frac{75}{28}\\ =\frac{-47}{28}\)
a, \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-2}{5}\right)\)
= \(\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)
b, \(0,75.1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\)
= \(\frac{3}{4}.\frac{16}{9}-\frac{7}{5}.\frac{-20}{21}\)
= \(\frac{4}{3}-\left(\frac{-4}{3}\right)=\frac{8}{3}\)
c, \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}+\frac{4}{19}+\frac{8}{23}\)
= \(\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)
= \(\left(-1\right)+1+\frac{4}{19}=0+\frac{4}{19}=\frac{4}{19}\)
d, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}-1\frac{3}{5}:25\%\)
=> \(\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}:25\%\)
= \(\frac{16}{5}.\frac{25}{8}-\frac{8}{5}.25:100\)
= 10 - 0,4 = 9,6
e, \(\left(\frac{7}{8}-\frac{1}{2}\right).2\frac{2}{3}-\frac{3}{7}.\left(2,5^2\right)\)
=> \(\frac{3}{8}.\frac{8}{3}-\frac{3}{7}.6,25\)
= \(1-\frac{75}{28}=\frac{-47}{28}\)
A= \(\left(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}\right)+\left(\frac{-6}{13}+\frac{1}{2}+\frac{4}{3}\right)\)
A= \(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}-\frac{6}{13}+\frac{1}{2}+\frac{4}{3}\)
A= \(\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{7}{13}+\frac{6}{13}\right)-\left(\frac{1}{3}-\frac{4}{3}\right)\)
A= \(1-1-\left(-1\right)\)
A= \(1\)
B= \(0,75+\frac{2}{5}+\left(\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\right)\)
B= \(\frac{3}{4}+\frac{2}{5}+\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\)
B= \(\left(\frac{3}{4}+\frac{5}{4}\right)+\left(\frac{2}{5}-\frac{7}{5}\right)+\frac{1}{9}\)
B= \(2-1+\frac{1}{9}\)
B= \(\frac{9}{9}+\frac{1}{9}\)
B= \(\frac{10}{9}\)
C= \(\left(\frac{-3}{2}.\frac{4}{3}\right).\left(\frac{-9}{2}\right)-\frac{1}{4}\)
C = \(-2.\left(\frac{-9}{2}\right)-\frac{1}{4}\)
C = \(9-\frac{1}{4}\)
C = \(\frac{36}{4}-\frac{1}{4}\)
C = \(\frac{35}{4}\)
D = \(\frac{5}{4}.\left(\frac{-7}{10}.\frac{5}{4}-\frac{7}{8}.\frac{7}{10}\right)\)
D = \(\frac{5}{4}.\left(\frac{-7}{8}-\frac{49}{80}\right)\)
D = \(\frac{-35}{32}-\frac{49}{64}\)
D = \(\frac{-70}{64}-\frac{49}{64}\)
D = \(\frac{-119}{64}\)
k mk nha ^_^