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a/ \(\frac{5}{6n}\)và \(\frac{7}{15}\)
=> MSC = \(6n\cdot15=90n\)
\(\Rightarrow\frac{5}{6n}=\frac{5\cdot15}{90n}=\frac{75}{90n}\)
\(\Rightarrow\frac{7}{15}=\frac{7\cdot6n}{90n}=\frac{42n}{90n}\)
b/ \(\frac{9x}{24}\)và \(\frac{12}{36}\)
=> MSC = 72
\(\Rightarrow\frac{9x}{24}=\frac{9x\cdot3}{72}=\frac{27x}{72}\)
\(\Rightarrow\frac{12}{36}=\frac{12\cdot2}{72}=\frac{24}{72}\)
a)MSC = 6n . 15 = 90n
5/6n = 5 . 15/60n . 15 = 75/90n
7/15 = 7 . 6n/15 . 6n =42n/90n
#Louis
\(\text{ Bài giải }\)
\(a,\text{ }\frac{7n}{15}\text{ và }\frac{20}{39}\)
\(BCNN\left(15,39\right)=195\)
\(\frac{7n}{15}=\frac{7n\cdot13}{15\cdot13}=\frac{91n}{195}\) \(\frac{20}{39}=\frac{20\cdot5}{39\cdot5}=\frac{100}{195}\)
\(b,\text{ }\frac{14}{41}\text{ và }\frac{17n}{54}\)
\(BCNN\left(41,54\right)=2214\)
\(\frac{14}{41}=\frac{14\cdot54}{41\cdot54}=\frac{756}{2214}\) \(\frac{17n}{54}=\frac{17n\cdot41}{54\cdot41}=\frac{697n}{2214}\)
\(\frac{2}{n}+\frac{2}{n+1}=\frac{2\left(n+1\right)}{n\left(n+1\right)}+\frac{2n}{n\left(n+1\right)}\)\(=\frac{2\left(n+1\right)+2n}{n\left(n+1\right)}=\frac{2n+2+2n}{n\left(n+1\right)}=\frac{4n+2}{n\left(n+1\right)}\)
\(\frac{1}{n\left(n+1\right)}+\frac{-2}{n+1}=\frac{1}{n\left(n+1\right)}+\frac{-2n}{n\left(n+1\right)}\)\(=\frac{1+\left(-2n\right)}{n\left(n+1\right)}=\frac{1-2n}{n\left(n+1\right)}\)
\(a,\)\(\frac{2}{n}\)và \(\frac{2}{n+1}\)
Có : \(\frac{2}{n}=\frac{2\left(n+1\right)}{n\left(n+1\right)}\)
\(\frac{2}{n+1}=\frac{2n}{n\left(n+1\right)}\)
Vậy ta có : \(\frac{2\left(n+1\right)}{n\left(n+1\right)}\)và \(\frac{2n}{n\left(n+1\right)}\)
\(b,\)\(\frac{1}{n\left(n+1\right)}\)và \(\frac{-2}{n+1}\)
Có : \(\frac{1}{n\left(n+1\right)}\)
\(\frac{-2}{n+1}=\frac{-2n}{n\left(n+1\right)}\)
Vậy ta có : \(\frac{1}{n\left(n+1\right)}\)và \(\frac{-2n}{n\left(n+1\right)}\)
