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\(y+y\times\frac{1}{3}\div\frac{2}{9}+y\div\frac{2}{7}=252\)
\(3y\times\frac{1}{3}\div\frac{2}{9}\div\frac{2}{7}=252\)
\(3y\times\frac{21}{4}=252\)
\(3y=252\div\frac{21}{4}\)
\(3y=48\)
\(y=48\div3\)
\(y=16\)
a, ( y+2436 ) : 12 = 407
( y+2436 )=407x12
( y+2436 )=4884
y=4884-2436
y=2448
\(\frac{2}{7}:y=\frac{10}{21}.\frac{9}{14}\)
\(\frac{2}{7}:y=\frac{15}{49}\)
\(y=\frac{2}{7}:\frac{15}{49}\)
\(y=\frac{2}{7}.\frac{49}{15}\)
\(y=\frac{14}{15}\)
\(y-\frac{1}{3}=\frac{10}{21}:\frac{15}{28}\)
\(y-\frac{1}{3}=\frac{10}{21}.\frac{28}{15}\)
\(y-\frac{1}{3}=\frac{8}{9}\)
\(y=\frac{8}{9}+\frac{1}{3}\)
\(y=\frac{8}{9}+\frac{3}{9}\)
\(y=\frac{11}{9}\)
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 !!!
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right).200x=4036\)
\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}.200x=4036\)
\(\Leftrightarrow\frac{1.2.3...99}{2.3.4....100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow2x=4036\)
\(\Leftrightarrow x=4036:2=2018\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{100}\right)\times200\times x=4036\)
=> \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}\times200\times x=4036\)
=> \(\frac{1\times2\times...\times99}{2\times3\times...\times100}\times200\times x=4036\)
\(\Rightarrow\frac{1}{100}\times200\times x=4036\)
\(\Rightarrow2\times x=4036\)
=> x = 2018
x\(\times\)(5+2+3)=100
x\(\times\)10=100
x=100\(\div\)10=10
Vậy x =10