\(\frac{-6}{-35}=\frac{6}{35}=\frac{6\cdot36}{35\cdot36}=\frac{216}{1260}\)

\(\frac{27}{-180}=\frac{-27}{180}=\frac{-27\cdot7}{180\cdot7}=\frac{-189}{1260}\)

\(\frac{-3}{-28}=\frac{3}{28}=\frac{3\cdot45}{28\cdot45}=\frac{135}{1260}\)

Dễ mà pạn

1) \(\frac{-6}{-35};\frac{27}{-180};\frac{-3}{-28}\)

Đầu tiên ta phải rút gọn và để thành mẫu tử

\(\frac{-6}{-35}=\frac{6}{35}\)( p. số tối giản )

\(\frac{27}{-180}=\frac{-27}{180}=\frac{\left(-27\right):9}{180:9}=\frac{-3}{20}\) ( p.số tối giản )

\(\frac{-3}{-28}=\frac{3}{28}\) ( p.số tối giản )

Bước 2 : Ta tìm BCNN

BCNN ( 35; 20; 28 ) = \(35=5.7;20=2^2.5;28=2^2.7\)

\(=2^2.5.7=140\)

Bước 3 : Tìm thừa số phụ của mỗi mẫu

Các thừa số phụ là : 4 ; 7 ; 5 .

Bước 3 : Nhân tử và mẫu cho thừa số phụ

\(\frac{6}{35}=\frac{6.4}{35.4}=\frac{24}{140}\)

\(\frac{-27}{180}=\frac{-3}{20}=\frac{\left(-3\right).7}{20.7}=\frac{-21}{140}\)

\(\frac{-3}{-28}=\frac{3}{28}=\frac{3.5}{28.5}=\frac{15}{140}\)

\(\frac{24}{140};\frac{-21}{140};\frac{15}{140}\)

\(\frac{3}{-20};\frac{-11}{-30};\frac{7}{15}\)

\(\frac{3}{-20}=\frac{3}{20}\)

\(\frac{-11}{-30}=\frac{11}{30}\)

\(BCNN\left(20;30;15\right)=60\)

\(60:20=3\)

\(60:30=2\)

\(60:15=4\)

\(\frac{-3}{20}=\frac{-3.3}{20.3}=\frac{-9}{60}\)

\(\frac{11}{30}=\frac{11.2}{30.2}=\frac{22}{60}\)

\(\frac{7}{15}=\frac{7.4}{15.4}=\frac{28}{60}\)

Câu b tương tự như vậy ấy 

Tk cho mk nha Thank you so much

a)\(\frac{3}{-20}=-\frac{9}{60}\);\(-\frac{11}{-30}=\frac{22}{60}\);\(\frac{7}{15}=\frac{28}{60}\)

b)\(-\frac{6}{-35}=\frac{216}{1260}\);\(\frac{27}{-180}=-\frac{189}{1260}\);\(-\frac{3}{-28}=\frac{135}{1260}\)

a)  = -17, =-41,  = 22                       b) = -41, = -153, = -31

Giải:

\(\dfrac{3}{2}+\dfrac{6}{8}+\dfrac{9}{28}+\dfrac{12}{77}+\dfrac{15}{176}\) 

\(=\dfrac{3}{1.2}+\dfrac{6}{2.4}+\dfrac{9}{4.7}+\dfrac{12}{7.11}+\dfrac{15}{11.16}\) 

\(=3.\left(\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}\right)\) 

\(=3.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\right)\) 

\(=3.\left(1-\dfrac{1}{16}\right)\) 

\(=3.\dfrac{15}{16}\) 

\(=\dfrac{45}{16}\)

Đặt A = \(\dfrac{3}{2}+\dfrac{6}{8}+\dfrac{9}{28}+\dfrac{12}{77}+\dfrac{15}{176}\)

= \(3\left(\dfrac{1}{1\times2}+\dfrac{2}{2\times4}+\dfrac{3}{4\times7}+\dfrac{4}{7\times11}+\dfrac{5}{11\times16}\right)\)

= \(3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\right)\)

= \(3\left(1-\dfrac{1}{16}\right)=3\times\dfrac{15}{16}=\dfrac{45}{16}\)

a: 27/-180=-27/180=-3/20=-21/140

-6/-35=6/35=24/120

-3/-28=3/28=15/140

b: \(\dfrac{3\cdot4+3\cdot7}{6\cdot5+9}=\dfrac{3\left(4+7\right)}{30+9}=\dfrac{11}{13}=\dfrac{2849}{13\cdot259}\)

\(\dfrac{6\cdot9-2\cdot17}{63\cdot6-119}=\dfrac{54-34}{259}=\dfrac{20}{259}=\dfrac{260}{259\cdot13}\)

-6/-35=6/35=24/140

3/28=15/140

27/-180=-3/20=-21/140