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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
\(\text{ Bài giải }\)
\(a,\text{ }\frac{7n}{15}\text{ và }\frac{20}{39}\)
\(BCNN\left(15,39\right)=195\)
\(\frac{7n}{15}=\frac{7n\cdot13}{15\cdot13}=\frac{91n}{195}\) \(\frac{20}{39}=\frac{20\cdot5}{39\cdot5}=\frac{100}{195}\)
\(b,\text{ }\frac{14}{41}\text{ và }\frac{17n}{54}\)
\(BCNN\left(41,54\right)=2214\)
\(\frac{14}{41}=\frac{14\cdot54}{41\cdot54}=\frac{756}{2214}\) \(\frac{17n}{54}=\frac{17n\cdot41}{54\cdot41}=\frac{697n}{2214}\)
\(\frac{2}{n}+\frac{2}{n+1}=\frac{2\left(n+1\right)}{n\left(n+1\right)}+\frac{2n}{n\left(n+1\right)}\)\(=\frac{2\left(n+1\right)+2n}{n\left(n+1\right)}=\frac{2n+2+2n}{n\left(n+1\right)}=\frac{4n+2}{n\left(n+1\right)}\)
\(\frac{1}{n\left(n+1\right)}+\frac{-2}{n+1}=\frac{1}{n\left(n+1\right)}+\frac{-2n}{n\left(n+1\right)}\)\(=\frac{1+\left(-2n\right)}{n\left(n+1\right)}=\frac{1-2n}{n\left(n+1\right)}\)
1. a) Ta có BCNN(12, 15) = 60 nên ta lấy mẫu chung của hai phân số là 60.
Thừa số phụ:
60:12 =5; 60:15=4
Ta được:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\)
\(\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\)
b) Ta có BCNN(7, 9, 12) = 252 nên ta lấy mẫu chung của ba phân số là 252.
Thừa số phụ:
252:7 = 36; 252:9 = 28; 252:12 = 21
Ta được:
\(\frac{2}{7} = \frac{{2.36}}{{7.36}} = \frac{{72}}{{252}}\)
\(\frac{4}{9} = \frac{{4.28}}{{9.28}} = \frac{{112}}{{252}}\)
\(\frac{7}{{12}} = \frac{{7.21}}{{12.21}} = \frac{{147}}{{252}}\)
2. a) Ta có BCNN(8, 24) = 24 nên:
\(\frac{3}{8} + \frac{5}{{24}} = \frac{{3.3}}{{8.3}} + \frac{5}{{24}} = \frac{9}{{24}} + \frac{5}{{24}} = \frac{{14}}{{24}} = \frac{7}{{12}}\)
b) Ta có BCNN(12, 16) = 48 nên:
\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).
a) Ta có: BCNN(16, 24) = 48
48 : 16 = 3; 48 : 24 = 2. Do đó:
\(\frac{3}{{16}} = \frac{{3.3}}{{16.3}} = \frac{9}{{48}}\)
\(\frac{5}{{24}} = \frac{{5.2}}{{24.2}} = \frac{{10}}{{48}}\).
b) Ta có: BCNN(20, 30, 15) = 60
60 : 20 = 3; 60 : 30 = 2; 60 : 15 = 4. Do đó:
\(\frac{3}{{20}} = \frac{{3.3}}{{20.3}} = \frac{9}{{60}}\)
\(\frac{{11}}{{30}} = \frac{{11.2}}{{30.2}} = \frac{{22}}{{60}}\)
\(\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\).
a) \(\frac{4}{9}\)và \(\frac{7}{15}\)
Ta có: \(9 = 3^2 ; 15 = 3.5\) nên \(BCNN (9,15) = 3^2. 5 = 45\). Do đó ta có thể chọn mẫu chung là 45.
\(\frac{4}{9}=\frac{4.5}{9.5}=\frac{20}{45}\)
\(\frac{7}{15}=\frac{7.3}{15.3}=\frac{21}{45}\)
b) \(\frac{5}{12}; \frac{7}{15}\) và \(\frac{4}{27}\)
Ta có: \(12=2^2.3\); \(15 = 3.5\) ; \(27=3^3\) nên BCNN(12, 15, 27) =\(2^2.3^3.5=540\). Do đó ta có thể chọn mẫu chung là 540.
\(\frac{5}{12}=\frac{5.45}{12.45}=\frac{225}{540}\)
\(\frac{7}{15}=\frac{7.36}{15.36}=\frac{252}{540}\)
\(\frac{4}{27}=\frac{4.20}{27.20}=\frac{80}{540}\)
a)
i.Ta có: BCNN(12, 30) = 60
60 : 12 = 5; 60 : 30 = 2. Do đó:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\) và \(\frac{7}{{30}} = \frac{{7.2}}{{30.2}} = \frac{{14}}{{60}}.\)
ii.Ta có: BCNN(2, 5, 8) = 40
40 : 2 = 20; 40 : 5 = 8; 40 : 8 = 5. Do đó:
\(\frac{1}{2} = \frac{{1.20}}{{2.20}} = \frac{{20}}{{40}}\)
\(\frac{3}{5} = \frac{{3.8}}{{5.8}} = \frac{{24}}{{40}}\)
\(\frac{5}{8} = \frac{{5.5}}{{8.5}} = \frac{{25}}{{40}}\).
b)
i.Ta có: BCNN(6, 8) = 24
24 : 6 = 4; 24: 8 = 3. Do đó
\(\begin{array}{l}\frac{1}{6} + \frac{5}{8} = \frac{{1.4}}{{6.4}} + \frac{{5.3}}{{8.3}}\\ = \frac{4}{{24}} + \frac{{15}}{{24}} = \frac{{19}}{{24}}.\end{array}\)
ii. Ta có: BCNN(24, 30) = 120
120: 24 = 5; 120: 30 = 4. Do đó:
\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)
\(a,\)\(\frac{2}{n}\)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 \(\frac{-2n}{n\left(n+1\right)}\)
a) Ta có: \(12 = 2^2 . 3; 15 = 3.5\)
\(BCNN(12, 15) = 2^2.3.5 = 60\) nên chọn mẫu chung là 60.
\(\begin{array}{l}\frac{9}{{12}} = \frac{{9.5}}{{12.5}} = \frac{{45}}{{60}}\\\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\end{array}\)
b) Ta có: \(10 = 2.5; 4 = 2^2; 14=2.7\)
\(BCNN(10, 4, 14) =2^2.5.7= 140\) nên chọn mẫu chung là 140.
\(\begin{array}{l}\frac{7}{{10}} = \frac{{7.14}}{{10.14}} = \frac{{98}}{{140}}\\\frac{3}{4} = \frac{{3.35}}{{4.35}} = \frac{{105}}{{140}}\\\frac{9}{{14}} = \frac{{9.10}}{{14.10}} = \frac{{90}}{{140}}\end{array}\)