\(a,=\dfrac{5x}{4y^3}\times\left(\dfrac{-20y}{x^4}\right)=\dfrac{-100xy}{4x^4y^3}=\dfrac{-25}{x^3y^2}\\ b,=\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x+4\right)}\times\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)

\(c,=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\times\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x+3\right)^2.\left(x^2+2x+4\right)}\)

a) \(\dfrac{5x}{4y^3}:\left(-\dfrac{x^4}{20y}\right)=\dfrac{5x}{4y^3}\cdot\left(-\dfrac{20y}{x^4}\right)=\dfrac{5\cdot-5}{y^2\cdot x^3}=\dfrac{-25}{x^3y^2}\)

b) \(\dfrac{x^2-16}{x+4}:\dfrac{2x-8}{x}=\left(x-4\right)\cdot\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)

c) \(\dfrac{2x+6}{x^3-8}:\dfrac{\left(x+3\right)^3}{2x-4}=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\cdot\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x^2+2x+4\right)\left(x+3\right)^2}\)

a) (-5x³ + 15x² + 18x) : (-5x)

= (-5x³) : (-5x) + 15x² : (-5x) + 18x : (-5x)

= [(-5): (-5)] . (x³ : x) + [15 : (-5)] . (x² : x) + [18 : (-5)]. (x : x)

=  x² – 3x - \(\dfrac{{18}}{5}\)

b) (-2x⁵ – 4x³ + 3x²) : 2x²

= (-2x⁵ : 2x²) + (-4x³ : 2x²) + (3x² : 2x²)

= [(-2) : 2] . (x⁵ : x²) + [(-4) : 2] . (x³ : x²) + (3 : 2) . (x² : x²)

= -x³ – 2x + \(\dfrac{3}{2}\)

`a, a/(a-3) - 3/(a+3) = (a(a+3) - 3(a-3))/(a^2-9)`

`= (a^2+9)/(a^2-9)`

`b, 1/(2x) + 2/x^2 = x/(2x^2) + 4/(2x^2) = (x+4)/(2x^2)`

`c, 4/(x^2-1) - 2/(x^2+x) = (4x)/(x(x-1)(x+1)) - (2(x-1))/(x(x+1)(x-1))`

`= (2x+2)/(x(x-1)(x+1)`

`= 2/(x(x-1))`

a) (0,5x⁵ + 3,2x³ – 2x² ) : 0,25x²

= 0,5x⁵ : 0,25x² + 3,2x³ : 0,25x² - 2x² : 0,25x²

= (0,5:0,25).(x⁵ : x²) + (3,2 : 0,25). (x³ : x² ) - (2 : 0,25). (x² : x²)

= 2x³ + 12,8x - 8

b) (0,5x⁵ + 3,2x³ – 2x² ) : 0,25x³

a) Ta có: 

\(A=-3\cdot7\cdot\left(-2\right)\cdot\left(-13\right)\)

\(A=-21\cdot26\)

\(A=-546\)

\(B=-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-4\right)\cdot5\)

\(B=2\cdot12\cdot5\)

\(B=2\cdot60\)

\(B=120\)

Mà: \(120>-546\)

\(\Rightarrow B>A\)

a) \(\dfrac{a-1}{a+1}+\dfrac{3-a}{a+1}\)

\(=\dfrac{a-1+3-a}{a+1}\)

\(=\dfrac{2}{a+1}\)

b) \(\dfrac{b}{a-b}+\dfrac{a}{b-a}\)

\(=\dfrac{b}{a-b}+\dfrac{-a}{a-b}\)

\(=\dfrac{b-a}{a-b}\)

\(=-1\)

c) \(\dfrac{\left(a+b\right)^2}{ab}-\dfrac{\left(a-b\right)^2}{ab}\)

\(=\dfrac{\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)+\left(a-b\right)\right]}{ab}\)

\(=\dfrac{4ab}{ab}\)

\(=4\)

`a, (a-1)/(a+1) + (3-a)/(a+1)`

`= (a-1+3-a)/(a+1)`

`=2/(a+1)`

`b, b/(a-b) + a/(b-a)`

`=  b/(a-b) - a/(a-b)`

`= (b-a)/(a-b)`

`c, (a+b)^2/(ab) -(a-b)^2/(ab)`

`=(a^2+2ab+b^2-a^2+2ab-b^2)/(ab)`

`= (4ab)/(ab)`

\(a,=\dfrac{4y.5x^3}{3x^2.2y^3}=\dfrac{20x^3y}{6x^2y^3}=\dfrac{10x}{3y^2}\\ b,=\dfrac{\left(x-1\right)^2.x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2.x.\left(x+1\right)}{\left(x-1\right)^2.\left(x+1\right)}=x\)

\(c,=\dfrac{x\left(2+x\right).3\left(x^3+1\right)}{\left(x^2-x+1\right).3.\left(x+2\right)}=\dfrac{3x.\left(x+2\right).\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-x+1\right).3\left(x+2\right)}=x\left(x+1\right)\)

a: \(=\dfrac{3b+4a}{6ab}\)

b: \(=\dfrac{x^2-2x+1-x^2-2x-1}{x^2-1}=\dfrac{-4x}{x^2-1}\)

c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)

d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)

e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)

a) 945 : 45 = 21

b) 3 121 : 51 = 61 (dư 10)

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:

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