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Dấu " . " k p dấu " , "
a) \(\frac{91}{1\cdot4}+\frac{91}{4\cdot7}+\frac{91}{7\cdot10}+...+\frac{91}{88\cdot99}\)
* Bài đúng k z ???
b) \(5\frac{1}{7}\left(3\frac{2}{3}+4\frac{1}{7}\right)=\frac{36}{7}\cdot\left(\frac{11}{3}+\frac{29}{7}\right)\)
\(=\frac{36}{7}\cdot\frac{164}{21}=\frac{1968}{49}\)
a/=(74-(-1937)1)
=74-(-1937)
=2011
b/=4/7+5/6:5-3/8*(-4)
=4/7+1/6-(-3/2)
=31/42-(-3/2)
=47/21
minh chi biet bay nhieu
\(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{3+\frac{3}{13}+\frac{3}{169}+\frac{3}{91}}{7+\frac{7}{13}+\frac{7}{169}+\frac{7}{91}}\)\(=\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{3.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{7.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\)
\(=\frac{12}{4}:\frac{3}{7}\)
\(=3.\frac{7}{3}=7\)
\(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{3+\frac{3}{13}+\frac{3}{169}+\frac{3}{91}}{7+\frac{7}{13}+\frac{7}{169}+\frac{7}{91}}\)
\(=\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{3\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{9}\right)}{7\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{9}\right)}\)
\(=3:\frac{3}{7}\)
\(=7\)
a) \(x-\frac{5}{7}=\frac{1}{9}\Rightarrow x=\frac{1}{9}+\frac{5}{7}\Rightarrow x=\frac{52}{63}\)
b) \(\frac{-3}{7}-x=\frac{4}{5}+\frac{-2}{3}\Rightarrow\frac{-3}{7}-x=\frac{2}{15}\Rightarrow x=\frac{-3}{7}-\frac{2}{15}\Rightarrow x=\frac{-59}{105}\)
c) \(x-\frac{1}{5}=\frac{2}{7}.\frac{-11}{5}\Rightarrow x-\frac{1}{5}=\frac{-22}{35}\Rightarrow x=\frac{-22}{35}+\frac{1}{5}\Rightarrow x=\frac{-3}{7}\)
d) \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Rightarrow\frac{x}{182}=\frac{-15}{91}\Rightarrow x=\frac{\left(-15\right).182}{91}\Rightarrow x=-30\)
Đấm zô đúng thì biết câu trả lời của mình hay lắm đó cứ thử đi
\(=\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{3.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{7.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\)
\(=\frac{12}{4}:\frac{3}{7}=3.\frac{7}{3}=7\)