A=\(\frac{10-1\frac{1}{6}.\frac{6}{7}}{21:\frac{11}{2}+5\frac{2}{11}}\)
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\(\frac{10-1\frac{1}{6}.\frac{6}{7}}{21:\frac{11}{2}+5\frac{2}{11}}\)
\(=\frac{10-\frac{7}{6}.\frac{6}{7}}{21.\frac{2}{11}+\frac{57}{11}}\)
\(=\frac{10-1}{\frac{42}{11}+\frac{57}{11}}\)
\(=\frac{9}{\frac{99}{11}}\)
\(=\frac{9}{9}=1\)
\(A=\frac{10-1\frac{1}{6}\times\frac{6}{7}}{21:\frac{11}{2}+5\frac{2}{11}}\)
\(A=\frac{10-\frac{7}{6}\times\frac{6}{7}}{21:\frac{11}{2}+\frac{57}{11}}\)
\(A=\frac{10-1}{\frac{42}{11}+\frac{57}{11}}\)
\(A=\frac{9}{9}=1\)
b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)
\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)
\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)
\(=\frac{22}{29}.\)
Chúc bạn học tốt!
a)\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)
\(=1.6:\frac{9^8.3-9^8.9}{9^8.\left(5+7\right)}\)
\(=6:\frac{9^8.\left(3-9\right)}{9^8.12}\)
\(=6:\frac{9^8.\left(-6\right)}{9^8.12}\)
\(=6:\left(-\frac{6}{12}\right)\)
\(=6:\left(-\frac{1}{2}\right)\)
\(=-12\)
b) 3/5 : ( -1/5-1/6)+3/5:(-1/3-16/15) ( mình chuyển về ps luôn )
=3/5: (-11/30) + 3/5 : (-7/5)
=3/5:[-11/30+(-7/5)]
=3/5:53/30
=18/53
c) (1/2-13/14):5/7-(-2/21+1/7):5/7
= -3/7:5/7-1/21:5/7
=(-3/7-1/21):5/7
=-10/21:5/7
=-2/3
câu b vá c mình làm tắt nha. chúc bạn học tốt
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Bài làm
\(\frac{10-1\frac{1}{6}\times\frac{6}{7}}{21:\frac{11}{2}+5\frac{2}{12}}\)
\(=\frac{10-\frac{7}{6}\times\frac{6}{7}}{21\times\frac{2}{11}+\frac{62}{12}}\)
\(=\frac{10-1}{\frac{42}{11}+\frac{62}{12}}\)
\(=9:(\frac{42}{11}+\frac{62}{12})\)
\(=9:\left(\frac{504}{132}+\frac{682}{132}\right)\)
\(=9:\frac{1186}{132}\)
\(=9:\frac{593}{66}\)
\(=9\times\frac{66}{593}\)
\(=\frac{594}{593}\)
# Hông bt có đúng k nx. #
a) Ta có: BCNN(15,10) = 30 nên ta chọn mẫu số chung là 30
\(\frac{11}{15}+\frac{9}{10}=\frac{22}{30}+\frac{27}{30}=\frac{49}{30}\)
b) Ta có: BCNN(6,9,12) = 36 nên ta chọn mẫu số chung là 36
\(\frac{5}{6} + \frac{7}{9} + \frac{{11}}{{12}} = \frac{{30}}{{36}} + \frac{{28}}{{36}} + \frac{{33}}{{36}} = \frac{{91}}{{36}}\)
c) Ta có: BCNN(24,21) = 168 nên ta chọn mẫu số chung là 168
\(\frac{7}{{24}} - \frac{2}{{21}} = \frac{{49}}{{168}} - \frac{{16}}{{168}} = \frac{{33}}{{168}}=\frac{11}{56}\)
d) Ta có: BCNN(36,24) = 72 nên ta chọn mẫu số chung là 72
\(\frac{{11}}{{36}} - \frac{7}{{24}} = \frac{{22}}{{72}} - \frac{{21}}{{72}} = \frac{1}{{72}}\)
Ta có :
\(P=\frac{\frac{6}{8}+\frac{6}{10}+\frac{6}{14}+\frac{6}{26}}{\frac{11}{4}+\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(\Rightarrow P=\frac{\frac{3}{4}+\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3}{11}\)
Vậy \(P=\frac{3}{11}\)
\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}=\frac{3}{11}\)
đề bài của bn sai nên mk sửa luôn nha
a) \(\frac{13}{26}-\frac{1}{3}-\frac{1}{2}+\frac{7}{21}\)
\(=\frac{1}{2}-\frac{1}{3}-\frac{1}{2}+\frac{1}{3}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)
\(=0+0\)
\(=0\)
b) \(\left(\frac{-5}{12}+\frac{6}{11}\right)+\left(\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\right)\)
\(=\frac{-5}{12}+\frac{6}{11}+\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\)
\(=\left(\frac{-5}{12}+\frac{5}{12}\right)+\left(\frac{7}{17}+\frac{5}{17}\right)+\frac{6}{11}\)
\(=0+\frac{12}{17}+\frac{6}{11}\)
\(=\frac{132}{187}+\frac{102}{187}\)
\(=\frac{234}{187}\)
c) \(\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{12}{10}\right)\)
\(=\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{6}{5}\right)\)
\(=\frac{13}{5}+\frac{7}{16}-\frac{11}{16}+\frac{6}{5}\)
\(=\left(\frac{13}{5}+\frac{6}{5}\right)+\left(\frac{7}{16}-\frac{11}{16}\right)\)
\(=\frac{19}{5}+\left(\frac{-4}{16}\right)\)
\(=\frac{19}{5}-\frac{1}{4}\)
\(=\frac{76}{20}-\frac{5}{20}\)
\(=\frac{71}{20}\)
d) \(-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{21}{30}-\frac{5}{11}\right)\)
\(=-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{7}{10}-\frac{5}{11}\right)\)
\(=-\frac{3}{10}+\frac{6}{11}-\frac{7}{10}+\frac{5}{11}\)
\(=
\left(-\frac{3}{10}-\frac{7}{10}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)\)
\(=\frac{-10}{10}+\frac{11}{11}\)
\(=-1+1\)
\(=0\)
\(A=\frac{10-1\frac{1}{6}.\frac{6}{7}}{21:\frac{11}{2}+5\frac{2}{11}}=\frac{10-\frac{7}{6}.\frac{6}{7}}{\frac{21.2}{11}+\frac{57}{11}}=\frac{10-1}{\frac{42}{11}+\frac{57}{11}}=\frac{9}{9}=1\)