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\(\frac{162}{67xy}=\frac{270}{335}\)
\(\frac{162}{67xy}=\frac{54}{67}\)
\(\frac{162}{67xy}=\frac{162}{201}\)
Sai đề bài à? k đi
a)\(\frac{7\cdot15:25\cdot102}{7\cdot4\cdot15\cdot3:25\cdot4\cdot102\cdot3}=\frac{1}{4\cdot3:4\cdot3}=\frac{1}{1}=1\)
b)\(\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}=\frac{2+3+4+5+6+7+8}{10}=\frac{44}{10}=\frac{22}{5}\)
1 \(A=\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times.........\times\left(1+\frac{1}{2016}\right)\times\left(1+\frac{1}{2017}\right)\)
\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)
\(A=\frac{2018}{2}=1009\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)
\(B=\frac{1}{3}-\frac{1}{45}\)
\(B=\frac{14}{45}\)
2 \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)
\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)
\(=\frac{2017}{2018}\times1\)
=\(\frac{2017}{2018}\)
bạn nào xem giải thế có đúng ko
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.\frac{35}{36}\)
\(=\frac{3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{7}{2.6}\)
\(=\frac{7}{12}\)
\(=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot5\cdot7}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot6\cdot6}\)
\(=\frac{1\cdot7}{2\cdot6}\)
\(=\frac{7}{12}\)
\(\frac{2}{7}:y=\frac{10}{21}.\frac{9}{14}\)
\(\frac{2}{7}:y=\frac{15}{49}\)
\(y=\frac{2}{7}:\frac{15}{49}\)
\(y=\frac{2}{7}.\frac{49}{15}\)
\(y=\frac{14}{15}\)
\(y-\frac{1}{3}=\frac{10}{21}:\frac{15}{28}\)
\(y-\frac{1}{3}=\frac{10}{21}.\frac{28}{15}\)
\(y-\frac{1}{3}=\frac{8}{9}\)
\(y=\frac{8}{9}+\frac{1}{3}\)
\(y=\frac{8}{9}+\frac{3}{9}\)
\(y=\frac{11}{9}\)
a) Ta có : \(y-\frac{1}{3}=\frac{10}{21}\div\frac{15}{28}\)
\(\Rightarrow\) \(y-\frac{1}{3}=\frac{8}{9}\)
\(\Rightarrow\) \(y\) \(=\frac{8}{9}+\frac{1}{3}\)
\(\Rightarrow\) \(y\) \(=\frac{11}{9}\)
Vậy \(y=\frac{11}{9}\)
b) Ta có : \(\frac{2}{7}\div y=\frac{10}{21}\times\frac{9}{14}\)
\(\Rightarrow\) \(\frac{2}{7}\div y=\frac{15}{49}\)
\(\Rightarrow\) \(y=\frac{2}{7}\div\frac{15}{49}\)
\(\Rightarrow\) \(y=\frac{14}{15}\)
Vậy \(y=\frac{14}{15}\)
Cbht !!!
a) \(\frac{16}{35}+\frac{8}{35}=\frac{24}{35}\)
b)\(\frac{160}{77}-\frac{28}{77}=\frac{132}{77}=\frac{12}{1}=12\)
c)\(\frac{72}{180}=\frac{18}{45}\)
d) \(\frac{90}{360}=\frac{1}{4}\)
\(\frac{16}{3}:\frac{8}{3}=\frac{16}{3}\times\frac{3}{8}=\frac{16}{8}=2\)
\(\frac{63}{35}:\frac{21}{5}=\frac{63}{35}.\frac{5}{21}=\frac{3}{7}.\frac{1}{1}=\frac{3}{7}\)
\(\frac{8}{14}\times\frac{7}{8}=\frac{7}{14}=\frac{1}{2}\)
\(\frac{19}{14}\times\frac{7}{15}=\frac{19}{2}\times\frac{1}{15}=\frac{19}{30}\)
\(a,\frac{16}{3}:\frac{8}{3}=\frac{16}{3}\times\frac{3}{8}=\frac{8\times2\times3}{3\times8}=2\)
\(b,\frac{63}{35}:\frac{21}{5}=\frac{63}{35}\times\frac{5}{21}=\frac{21\times3\times5}{5\times7\times21}=\frac{3}{7}\)
\(c,\frac{8}{14}\times\frac{7}{8}=\frac{8\times7}{7\times2\times8}=\frac{1}{2}\)
\(d,\frac{19}{14}\times\frac{7}{15}=\frac{19\times7}{7\times2\times15}=\frac{19}{30}\)
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đó đâu phải toán lớp 4