\(\frac{5}{12}và\frac{2}{3}=\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a) $\frac{5}{3} = \frac{{5 \times 4}}{{3 \times 4}} = \frac{{20}}{{12}}$
Quy đồng mẫu số hai phân số $\frac{5}{3}$ và $\frac{{28}}{{12}}$ được $\frac{{20}}{{12}}$ và $\frac{{28}}{{12}}$
b) $\frac{2}{9} = \frac{{2 \times 9}}{{9 \times 9}} = \frac{{18}}{{81}}$
Quy đồng mẫu số hai phân số $\frac{2}{9}$ và $\frac{{17}}{{81}}$ được $\frac{{18}}{{81}}$ và $\frac{{17}}{{81}}$
c) $\frac{5}{8} = \frac{{5 \times 3}}{{8 \times 3}} = \frac{{15}}{{24}}$
Quy đồng mẫu số hai phân số $\frac{9}{{24}}$và $\frac{5}{8}$ được $\frac{9}{{24}}$và $\frac{{15}}{{24}}$
a) \(\dfrac{28}{12}=\dfrac{28:4}{12:4}=\dfrac{7}{3}\)
b) \(\dfrac{2}{9}=\dfrac{2\times9}{9\times9}=\dfrac{18}{81}\)
c) \(\dfrac{9}{24}=\dfrac{9:3}{24:3}=\dfrac{3}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
= 4/5 + 1/5 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
= (4/5 + 1/5) + (1/2 + 1/2) + (1/2 + 1/2) + 1/2
= 1 + 1 + 1 + 1/2
= 3 + 1/2
= 6/2 + 1/2 (Hoặc ra luôn là hỗn số 3 và 1/2)
= 7/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đúng
b) Đúng
c) Sai vì phân số $\frac{{12}}{7}$ có mẫu số là 7
d) Đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}+\frac{\frac{3}{2}+1+\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}+\frac{5}{4}}\)
\(=\frac{3.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{3}{5}\)
\(=\frac{6}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}=\frac{5}{3.4}+\frac{5}{4.5}+\frac{5}{5.6}+...+\frac{5}{99.100}\)
\(5\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(5\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{97}{60}\)
So sánh
\(\frac{2}{3}=\frac{8}{12}\) > \(\frac{5}{12}\)
Vì 8 > 5 nên \(\frac{8}{12}>\frac{5}{12}\)
\(\Rightarrow\frac{2}{3}>\frac{5}{12}\)
quy đồng hả :
\(\frac{2}{3}\) = \(\frac{8}{12}\) giữ ngyên phân số \(\frac{5}{12}\)
ko có đề bài nha bạn !