\(\left(\dfrac{1}{x}+x-2\right):\left(\dfrac{1}{x^2-x}+1-\dfrac{3}{x-1}\right)\)

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

\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{x\left(x-1\right)}{x^2-4x+1}=\dfrac{\left(x-1\right)^3}{x^2-4x+1}\)

\(\dfrac{3}{7}\times\dfrac{7}{9}\times\dfrac{1}{2}\)

\(=\dfrac{3\times7\times1}{7\times9\times2}\)

\(=\dfrac{21}{126}\)

\(=\dfrac{1}{6}\)

\(\dfrac{5}{8}\times4\times\dfrac{1}{2}\\ =\dfrac{5}{8}\times\dfrac{4}{1}\times\dfrac{1}{2}\\ =\dfrac{5\times4\times1}{8\times1\times2}\\ =\dfrac{20}{16}\\ =\dfrac{5}{4}\)

\(4\times\dfrac{1}{24}\times3\\ =\dfrac{4}{1}\times\dfrac{1}{24}\times\dfrac{3}{1}\\ =\dfrac{4\times1\times3}{1\times24\times1}\\ =\dfrac{12}{24}\\ =\dfrac{1}{2}\)

Bài 5: 

a. 1 - 2y + y²

= (1 - y)²

b. (x + 1)² - 25

= (x + 1)² - 5²

= (x + 1 - 5)(x + 1 + 5)

= (x - 4)(x + 6)

c. 1 - 4x²

= 1² - (2x)²

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

d. 8 - 27x³

= 2³ - (3x)³

= (2 - 3x)(4 + 6x + 9x²)

e. (đề hơi khó hiểu ''x3'' !?)

g. x³ + 8y³

= (x + 2y)(x² - 2xy + y²)

a: \(=\dfrac{x^2-2x+1}{x}:\dfrac{x-1-3x^2+3x-3}{\left(x-1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{-2x^2+4x-4}\)

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

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

\(=\dfrac{x^3-3x^2+3x+1+2x^2-4x+1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)

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

a) \(\dfrac{1}{2}x(6x - 4) = \dfrac{1}{2}x.6x + \dfrac{1}{2}x.( - 4) = 3{x^2} - 2x\).

b) \(\begin{array}{l} - {x^2}(\dfrac{1}{3}{x^2} - x - \dfrac{1}{4}) =  - {x^2}.\dfrac{1}{3}{x^2} +  - {x^2}. - x +  - {x^2}. - \dfrac{1}{4}\\ =  - \dfrac{1}{3}{x^4} + {x^3} + \dfrac{1}{4}{x^2}\end{array}\)

Bài 2:

a) \(=x^2-36y^2\)

b) \(=x^3-8\)

Bài 3:

a) \(=x^2+2x+1-x^2+2x-1-3x^2+3=-3x^2+4x+3\)

b) \(=6\left(x-1\right)\left(x+1\right)=6x^2-6\)

\(\dfrac{2x+4}{x^3-1}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2x+4-2x^2-2x-2+x^2-x+2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

`a, 2/(x+1)` hay `2/(x-1)` cậu nhỉ?

`b,`

\(\dfrac{x-1}{x^2-5x+6}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x-2\right)^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x-1-\left(x^2-6x+9\right)+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{x-1-x^2+6x-9+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3x-6}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3}{x-3}\)