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28 tháng 6 2018

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-hoi chi do

28 tháng 6 2018

Mình nghĩ đề phải nhứ thế này

\(\frac{6-\frac{6}{13}+\frac{6}{19}-\frac{6}{27}}{5-\frac{5}{13}+\frac{5}{19}-\frac{5}{27}}\)

\(\frac{6.\left(1-\frac{1}{13}+\frac{1}{19}-\frac{1}{27}\right)}{5.\left(1-\frac{1}{13}+\frac{1}{19}-\frac{1}{27}\right)}\)

\(\frac{6}{5}\)

\(1\frac{1}{5}\)

16 tháng 7 2017

1/ \(\frac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)

\(=\frac{5^{20}.3^{29}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}=\frac{3^{29}.5^{18}.\left(25-9\right)}{3^{29}.5^{18}.\left(7-5\right)}=\frac{16}{2}=8\)

CÁC BÀI CÒN LẠI TƯƠNG TỰ HẾT NHÉ E

15 tháng 3 2016

\(\left(-1\frac{1}{6}\right)\left(\frac{1-\frac{3}{5}+\frac{3}{11}-\frac{3}{13}}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right)\left(\frac{4-\frac{4}{17}+\frac{4}{19}-\frac{4}{2013}}{5-\frac{5}{7}+\frac{5}{19}-\frac{5}{2013}}\right)\)

\(=-\frac{7}{6}.\left(\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}\right)}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right):\left(\frac{4.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}{5.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}\right)\)

\(=-\frac{7}{6}.3:\frac{4}{5}=-\frac{7}{2}.\frac{5}{4}=-\frac{35}{8}\)

30 tháng 4 2019

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)

\(=3+6+\left(\frac{14}{19}+\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)

\(=9+1+1+\frac{13}{17}\)

\(=11+\frac{13}{17}\)

\(=\frac{200}{17}\)

3 tháng 5 2019

Giúp Mik ik mai nộp oy

31 tháng 8 2017

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)

\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)

\(=\frac{1454}{323}+\frac{35}{43}+6\)

\(=5,...+6\)

\(=11,...\)

3 tháng 7 2018

\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)

\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)

\(=\sqrt{3}-2\) 

\(VayA=\sqrt{3}-2\)

a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)

\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)

\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)

b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)

\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)

\(\simeq40.39\)