a, \(\frac{5}{6};\frac{11}{12}\)

Giữ nguyên phân số \(\frac{11}{12}\)

Quy đồng: \(\frac{5}{6}=\frac{5\cdot2}{6\cdot2}=\frac{10}{12}\)

b, \(\frac{3}{5};-2\)

Giữ nguyên phân số \(\frac{3}{5}\)

Quy đồng: \(-2=\frac{-2\cdot5}{5}=-\frac{10}{5}\)

c, \(-\frac{1}{4};\frac{3}{8};\frac{7}{12}\)

Quy đồng: \(-\frac{1}{4}=\frac{-1\cdot6}{4\cdot6}=-\frac{6}{24};\frac{3}{8}=\frac{3\cdot3}{8\cdot3}=\frac{9}{24};\frac{7}{12}=\frac{7\cdot2}{12\cdot2}=\frac{14}{24}\)

d, \(-\frac{8}{9};\frac{17}{18};\frac{5}{-6}\)

Giữ nguyên phân số \(\frac{17}{18}\)

Quy đồng: \(-\frac{8}{9}=\frac{-8\cdot2}{9\cdot2}=-\frac{16}{18};\frac{5}{-6}=\frac{5\cdot3}{-6\cdot3}=\frac{15}{-18}=-\frac{15}{18}\)

giúp với ạ

giải dc nhưng mà hoi lâu

a)\(\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{45}{63}+\dfrac{28}{63}=\dfrac{73}{63}\) ; \(\dfrac{9}{11}+\dfrac{3}{8}=\dfrac{72}{88}+\dfrac{33}{88}=\dfrac{105}{88}\)

\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\); \(\dfrac{16}{25}-\dfrac{2}{5}=\dfrac{16}{25}-\dfrac{10}{25}=\dfrac{6}{25}\)

yutyugubhujyikiu

a) $\frac{1}{6}:\frac{3}{7} = \frac{1}{6} \times \frac{7}{3} = \frac{7}{{18}}$  

b) $\frac{5}{{12}}:\frac{1}{4} = \frac{5}{{12}} \times \frac{4}{1} = \frac{{5 \times 4}}{{12 \times 1}} = \frac{{5 \times 4}}{{4 \times 3 \times 1}} = \frac{5}{3}$

c) $\frac{4}{{15}}:\frac{8}{3} = \frac{4}{{15}} \times \frac{3}{8} = \frac{{4 \times 3}}{{15 \times 8}} = \frac{{4 \times 3}}{{5 \times 3 \times 4 \times 2}} = \frac{1}{{10}}$          

d) $\frac{{18}}{5}:\frac{9}{{10}} = \frac{{18}}{5} \times \frac{{10}}{9} = \frac{{18 \times 10}}{{5 \times 9}} = \frac{{9 \times 2 \times 5 \times 2}}{{5 \times 9}} = 4$

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:

\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).

Bài 1:

a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)

\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)

\(=\frac{-3}{5}\)

b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)

\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)

\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)

c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)

\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)

\(=3+2=5\)

d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)

\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)

\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)

e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)

\(=-1+1+\frac{-7}{9}\)

\(=-\frac{7}{9}\)

f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)

\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)

\(=1+\frac{-1}{7}=\frac{7}{7}+\frac{-1}{7}=\frac{6}{7}\)

\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)

\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)

\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)

\(\Rightarrow-1\le x< 6\)

\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)

Bài b tương tự

bạn ơi bạn giải câu b được ko. mk ko biết làm câu b