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a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
\(\dfrac{2}{5}+\dfrac{4}{9}=\dfrac{18}{45}+\dfrac{20}{45}=\dfrac{18+20}{45}=\dfrac{38}{45}\)
a) \(\dfrac{4}{5}\cdot\dfrac{5}{8}:\dfrac{4}{5}=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\cdot\dfrac{5}{8}=1\cdot\dfrac{5}{8}=\dfrac{5}{8}\)
b) \(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)=\dfrac{5}{6}+\left(\dfrac{1}{2}\cdot\dfrac{2}{3}+\dfrac{4}{5}\right)=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)=\dfrac{17}{15}+\dfrac{5}{6}=\dfrac{59}{30}\)
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
a; A = \(\dfrac{4026\times2014+4030}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2014+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2013\times2+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-4026+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2011\right)}{2013\times2016-2011}\)
A = 2
\(a,2\dfrac{2}{5}:y\times1\dfrac{3}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y\times\dfrac{7}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y=\dfrac{7}{8}:\dfrac{7}{4}\\ \dfrac{12}{5}:y=\dfrac{1}{2}\\ y=\dfrac{12}{5}:\dfrac{1}{2}=\dfrac{24}{5}\\ b,3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\\ \dfrac{17}{5}:y:\dfrac{5}{4}=\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{5}:\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{13}\\ y=\dfrac{17}{13}\times\dfrac{5}{4}=\dfrac{85}{52}\)
\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)
Câu đầu em xem lại đề bài sao có hai dấu bằng.
Câu 2:
\(\dfrac{3}{2}\) \(\times\)y - \(\dfrac{3}{4}\) \(\times\)y + y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{3}{2}\) - \(\dfrac{3}{4}\) + 1) = \(\dfrac{4}{5}\)
y \(\times\) (\(\dfrac{6}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{4}{4}\)) = \(\dfrac{4}{5}\)
y \(\times\) \(\dfrac{7}{4}\) = \(\dfrac{4}{5}\)
y = \(\dfrac{4}{5}\): \(\dfrac{7}{4}\)
y = \(\dfrac{16}{35}\)
Bài 4:
\(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)
\(\Rightarrow y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)
\(\Rightarrow y:\dfrac{2}{5}=\dfrac{25}{16}\)
\(\Rightarrow y=\dfrac{2}{5}\cdot\dfrac{25}{16}\)
\(\Rightarrow y=\dfrac{5}{8}\)
________________
\(456+y:87=23987\)
\(\Rightarrow y:87=23987-456\)
\(\Rightarrow y:87=23531\)
\(\Rightarrow y=23531\cdot87\)
\(\Rightarrow y=2047197\)
a)\(\dfrac{4}{5}\times\dfrac{5}{8}:\dfrac{4}{5}\)
\(=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\times\dfrac{5}{8}\)
\(=1\times\dfrac{5}{8}=\dfrac{5}{8}\)
b)\(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)\)
\(=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)\)
\(=\dfrac{5}{6}+\dfrac{17}{15}\)
\(=\dfrac{59}{30}\)
Bài 2:
a) \(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)
\(y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)
\(y:\dfrac{2}{5}=\dfrac{25}{16}\)
\(y=\dfrac{25}{16}\times\dfrac{2}{5}\)
\(y=\dfrac{5}{8}\)
b)\(456+y:87=23987\)
\(y:87=23987-456\)
\(y:87=23531\)
\(y=23531\times87\)
\(y=2047197\)