Ta có: \(\dfrac{5}{7} = \dfrac{{5.4}}{{7.4}} = \dfrac{{20}}{{28}}\) và \(\dfrac{{ - 3}}{4} = \dfrac{{ - 3.7}}{{4.7}} = \dfrac{{ - 21}}{{28}}\)

Như vậy, \(\dfrac{{20}}{{28}} + \dfrac{{ - 21}}{{28}} = \dfrac{{20 + \left( { - 21} \right)}}{{28}} =  \dfrac{-1}{{28}}\)

a) \(\left\{{}\begin{matrix}\dfrac{1}{5}=\dfrac{1.6}{5.6}=\dfrac{6}{30}\\\dfrac{1}{6}=\dfrac{1.5}{6.5}=\dfrac{5}{30}\\\dfrac{2}{15}=\dfrac{2.2}{15.2}=\dfrac{4}{30}\\\dfrac{1}{10}=\dfrac{1.3}{10.3}=\dfrac{3}{30}\end{matrix}\right.\)

Quy luật: Tử số của mỗi phân số cách nhau \(1\) đơn vị, cùng chung mẫu số là \(30\).

Phân số tiếp theo: \(\dfrac{2}{30}=\dfrac{1}{15}\)

b) \(\left\{{}\begin{matrix}\dfrac{1}{9}=\dfrac{1.5}{9.5}=\dfrac{5}{45}\\\dfrac{1}{15}=\dfrac{1.3}{15.3}=\dfrac{3}{45}\end{matrix}\right.\)

Quy luật: Tử số của mỗi phân số cách nhau \(1\) đơn vị, cùng chung mẫu số là \(45\).

Phân số tiếp theo: \(\dfrac{1}{45}\)

a, \(\dfrac{6}{30};\dfrac{5}{30};\dfrac{4}{30};\dfrac{3}{30};\dfrac{2}{30}\)

b,\(\dfrac{5}{45};\dfrac{4}{45};\dfrac{3}{45};\dfrac{2}{45};\dfrac{1}{45}\)

\(a,\dfrac{6}{30};\dfrac{5}{30};\dfrac{4}{30};\dfrac{3}{30};\dfrac{2}{30}\)

\(b,\dfrac{5}{45};\dfrac{4}{45};\dfrac{3}{45};\dfrac{2}{45};\dfrac{1}{45}\)

+ Quy đồng mẫu các phân số: \(\dfrac{3}{4}\) và \(\dfrac{5}{6}\):

\(BCNN\left( {6,4} \right) = 12\)

Thừa số phụ: \(12:4 = 3; 12:6=2\)

Ta có: \(\dfrac{3}{4} = \dfrac{{3.3}}{{4.3}} = \dfrac{9}{{12}}\\\dfrac{5}{6} = \dfrac{{5.2}}{{6.2}} = \dfrac{{10}}{{12}}\)

+ So sánh hai phân số cùng mẫu:

Vì 9 < 10 nên \(\dfrac{9}{{12}} < \dfrac{{10}}{{12}}\) nên \(\dfrac{3}{4} < \dfrac{5}{6}\).

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

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

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

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

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

giúp tớ

Ta có: 
\(MSC:36\)
\(\dfrac{8}{9}=\dfrac{8\times4}{9\times4}=\dfrac{32}{36}\)
\(\dfrac{2}{4}=\dfrac{2\times9}{4\times9}=\dfrac{18}{36}\)

`MSC: 36`

\(\dfrac{8}{9}=\dfrac{8\times4}{9\times4}=\dfrac{32}{36}\)

\(\dfrac{2}{4}=\dfrac{2\times9}{4\times9}=\dfrac{18}{36}\)

\(BCNN\left( {4,9,3} \right) = 36\)

Thừa số phụ: \(36:4 = 9;36:9 = 4;36:3 = 12\)

Ta có :

\(\begin{array}{l}\dfrac{{ - 3}}{4} = \dfrac{{ - 3.9}}{{4.9}} = \dfrac{{ - 27}}{{36}}\\\dfrac{5}{9} = \dfrac{{5.4}}{{9.4}} = \dfrac{{20}}{{36}}\\\dfrac{2}{3} = \dfrac{{2.12}}{{3.12}} = \dfrac{{24}}{{36}}\end{array}\)

\(\dfrac{-3}{4}=\dfrac{-3.9}{4.9}=\dfrac{-27}{36}\)

\(\dfrac{5}{9}=\dfrac{5.4}{9.4}=\dfrac{20}{36}\)

\(\dfrac{2}{3}=\dfrac{2.12}{3.12}=\dfrac{24}{36}\)

BCNN (4,9,3) = 36

\(\dfrac{-3}{4}=\dfrac{-3.9}{4.9}=\dfrac{-27}{36}\)

\(\dfrac{5}{9}=\dfrac{5.4}{9.4}=\dfrac{20}{36}\)

\(\dfrac{2}{3}=\dfrac{2.12}{3.12}=\dfrac{24}{36}\)

a) \(\dfrac{2}{36}=\dfrac{1}{18}\)

\(\dfrac{8}{12}=\dfrac{2}{3}\)

b) \(\dfrac{10}{25}=\dfrac{2}{5}\)

\(\dfrac{14}{40}=\dfrac{7}{20}\)

Các bn viết chi tiết ra giúp mk nha  Thanks 

\(1,\\ a,=xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\\ b,=\left(x-7\right)\left(x+7\right):\left(x-7\right)=x+7\\ 2,\dfrac{1}{a^2}-ab=\dfrac{1-a^3b}{a^2};\dfrac{1}{a^2}\text{ giữ nguyên}\\ 3,=\dfrac{-7}{t}\\ 4,=\dfrac{1-x+1-y}{x-y}=\dfrac{2-x-y}{x-y}\)

Bài 1:

\(a,\left(16x^3y^2-24x^2y^3+20x^4\right):16x^2=16x^2\left(xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\right):16x^2=xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\)

\(b,\left(x^2-49\right):\left(x-7\right)=\left[\left(x-7\right)\left(x+7\right)\right]:\left(x-7\right)=x+7\)

Bài 2:

\(\dfrac{1}{a^2}-ab=\dfrac{1-a^2b}{a^2}\)

\(\dfrac{1}{a^2}\)

Bài 3:

\(\dfrac{7\left(t-z\right)}{t\left(z-t\right)}=\dfrac{-7\left(z-t\right)}{t\left(z-t\right)}=\dfrac{-7}{t}\)

Bài 4:

\(\dfrac{x-1}{y-x}+\dfrac{1-y}{x-y}=\dfrac{x-1}{y-x}-\dfrac{1-y}{y-x}=\dfrac{x-1-1+y}{y-x}=\dfrac{x+y-2}{y-x}\)