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\(\frac{\frac{4}{11}-\frac{12}{31}+\frac{16}{59}}{\frac{3}{11}-\frac{9}{31}+\frac{12}{59}}=\frac{4.\left(\frac{1}{11}-\frac{3}{31}+\frac{4}{59}\right)}{3.\left(\frac{1}{11}-\frac{3}{31}+\frac{4}{59}\right)}=\frac{4}{3}\)( vì \(\frac{1}{11}-\frac{3}{31}+\frac{4}{59}\ne0\))
Bài làm:
Ta có: \(\frac{\frac{4}{11}-\frac{12}{31}+\frac{16}{59}}{\frac{3}{11}-\frac{9}{31}+\frac{12}{95}}=\frac{4\left(\frac{1}{11}-\frac{3}{31}+\frac{4}{59}\right)}{3\left(\frac{1}{11}-\frac{3}{31}+\frac{4}{59}\right)}=\frac{4}{3}\)
\(a,(\frac{1}{4}+\frac{-5}{13})+(\frac{2}{11}+\frac{-8}{13}+\frac{3}{4})\)
\(= (\frac{1}{4} + \frac{3}{4}) + (\frac{-5}{13} + \frac{-8}{13}) + \frac{2}{11}\)
\(= \frac{4}{4} + \frac{-13}{13} + \frac{2}{11}\)
\(=1+(-1)+\frac{2}{11}=\frac{2}{11}\)
\(b,(\frac{21}{31}+\frac{-16}{7})+(\frac{44}{53}+\frac{10}{31})+\frac{9}{53}\)
\(= (\frac{21}{31} + \frac{10}{31}) + (\frac{44}{53} + \frac{9}{53}) + \frac{-16}{7}\)
\(=\frac{31}{31}+\frac{53}{53}+\frac{-16}{7}=1+1-\frac{16}{7}\)
\(=2-\frac{16}{7}=\frac{14}{7}-\frac{16}{7}=-\frac27\)
\(c,\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)
\(= (\frac{-5}{7} + \frac{-2}{7}) + (\frac{3}{4} + \frac{1}{4}) + \frac{-1}{5}\)
\(= \frac{-7}{7} + \frac{4}{4} + \frac{-1}{5}\)
\(=-1+1-\frac{1}{5}=0-\frac15=-\frac15\)
\(\frac{-3}{31} + \frac{-6}{17} + \frac{1}{25} + \frac{-28}{31} + \frac{-11}{17} + \frac{-1}{5}\)
\(= (\frac{-3}{31} + \frac{-28}{31}) + (\frac{-6}{17} + \frac{-11}{17}) + \frac{1}{25} + \frac{-1}{5}\)
\(= \frac{-31}{31} + \frac{-17}{17} + \frac{1}{25} - \frac{5}{25}\)
\(= -1 + (-1) + \frac{-4}{25}\)
\(=-2-\frac{4}{25}=-\frac{50}{25}-\frac{4}{25}=-\frac{54}{25}\)
1/ ĐÁP ÁN:
\(\frac{-9}{33}=\frac{3}{-11}\); \(\frac{15}{9}=\frac{5}{3}\); \(\frac{-12}{19}=\frac{60}{-95}\)
2/ ĐÁP ÁN:
\(\frac{-7}{20}=\frac{3}{-18}=\frac{-9}{54}\ne\frac{12}{18}=\frac{-10}{-15}\ne\frac{14}{20}\)
3/ ĐÁP ÁN:
\(\frac{2}{3}=\frac{40}{60}\); \(\frac{3}{4}=\frac{45}{60}\); \(\frac{4}{5}=\frac{48}{60}\); \(\frac{5}{6}=\frac{50}{60}\)
a) \(\left(\frac{1}{4}+-\frac{5}{13}\right)+\left(\frac{2}{11}+-\frac{8}{13}+\frac{3}{4}\right)\)
\(=\frac{2}{11}+\left(\frac{1}{4}+\frac{3}{4}\right)+\left(-\frac{5}{13}+-\frac{8}{13}\right)\)
\(=\frac{2}{11}+1+\left(-1\right)\)
\(=\frac{2}{11}+0\)
\(=\frac{2}{11}\)
b) \(\left(\frac{21}{31}+-\frac{16}{7}\right)+\left(\frac{44}{53}+\frac{10}{31}\right)+\frac{9}{53}\)
\(=-\frac{16}{7}+\left(\frac{21}{31}+\frac{10}{31}\right)+\left(\frac{44}{53}+\frac{9}{53}\right)\)
\(=-\frac{16}{7}+1+1\)
\(=-\frac{16}{7}+2\)
=\(-\frac{2}{7}\)
c) \(\frac{\frac{9}{45}-\frac{4}{13}-\frac{1}{3}}{\frac{3}{13}-\frac{1}{15}+\frac{2}{3}}\)
\(=-\frac{43}{81}\)
bài 1b)
\(8\frac{1}{14}-6\frac37\)
C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)
C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)
\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)
bài 1 c)\(7-3\frac67\)
C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)
C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)
a) (1/4+3/4)+(-5/13+-8/13)+2/11
=4/4+-13/13+2/11
=1+(-1)+2/11
=2/11
k cho mk nha
a, Ta có : \(\frac{xy^2}{yz}=\frac{xyy}{yz}=\frac{xy}{z}.\frac{y}{y}=\frac{xy}{z}.1=\frac{xy}{z}\)
b, Ta có : \(\frac{7x-21}{14x-42}=\frac{7\left(x-3\right)}{14\left(x-3\right)}=\frac{7}{14}=\frac{1}{2}\)
c, Ta có : \(\frac{\overline{ab}}{abab}=\frac{10a+b}{1000a+100b+10a+b}=\frac{10a+b}{100\left(10a+b\right)+1\left(10a+b\right)}\)
\(=\frac{10a+b}{\left(100+1\right)\left(10a+b\right)}=\frac{1}{101}\)
d, Ta có : \(\frac{\frac{4}{11}-\frac{12}{31}+\frac{16}{59}}{\frac{3}{11}-\frac{9}{31}+\frac{12}{59}}=\frac{4\left(\frac{1}{11}-\frac{3}{31}+\frac{4}{59}\right)}{3\left(\frac{1}{11}-\frac{3}{31}+\frac{4}{59}\right)}=\frac{4}{3}\)