\(=\dfrac{x^2-4y}{xy}\cdot\dfrac{x^2}{x-y}=\dfrac{x\left(x^2-4y\right)}{y\left(x-y\right)}\)

a) 2(x-1)² - 4(x+3)² + 2x(x-5)

= 2(x² -2x +1)- 4(x² + 6x +9) + 2x² -10x

= 2x² - 4x + 2 -4x² - 24x - 36 + 2x² - 10x

= (2x² + 2x² - 4x²) - (4x + 24x+10x) +(2-36)

= -38x-34

b) 2(2x+5)²  -3(4x+1)(1-4x)

= 2(4x² + 20x + 25) + 3(4x+1)(4x-1)

= 8x² +40x + 50 + 3(16x² -1)

= 8x² + 40x + 50 + 48x² - 3

=56x² +40x + 47

a, \(2\left(x-1\right)^2-4\left(x+3\right)^2+2x\left(x-5\right)\)

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

\(=2x^2-4x+2-4x^2-24x-36+2x^2-10=-28x-44\)

b, \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)

\(=2\left(4x^2+20x+25\right)-3\left(1-16x^2\right)\)

\(=8x^2+40x+50-3+48x^2=56x^2+40x+47\)

mình sửa dòng cuối phần a nhé 

\(=2x^2-4x+2-4x^2-24x-36+2x^2-10x=-38x-34\)

Ta có : ( x - 1 ) ( x + 1 ) ( x + 2 )

= ( x² - 1 ) ( x + 2 )

= x³ + 2x² - x - 2

`th1:`

`(x+1)(x^2-x+1):(x-3)(x^2+3x+9)`

`=(x^3+1^3):(x^3-3^3)`

`=(x^3+1):(x^3-27)`

`=(x^3+1)/(x^3-27)`

`=(x^3-27+28)/(x^3-27)`

`=1+28/(x^3-27)`

`**th2:`

`(x+1)(x^2-x+1)`

`=x^3+1^3=x^3+1`

`(x-3)(x^2+3x+9)`

`=x^3-3^3=x^3-27`

\(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)

\(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)

\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=4x\left(x^2-9\right)-x^3+27\)

\(=4x^3-36x-x^3+27\)

\(=3x^3-36x+27\)

\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)

\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)

\(=\left(x+6\right).0\)

\(=0\)

Mình làm phần tiếp của bạn Phạm Tuấn Đạt nha .

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

\(=x^2+12x+36-2x^2-12x+x^2-36\)

\(=0\)

2. 

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

\(4x^2-4x+1-\left(4x^2-12x+x-3\right)=-3\)

\(4x^2-4x+1-4x^2+12x-x+3+3=0\)

\(7x+7=0\)

\(x=-1\)

\(\dfrac{2}{67}-\left(\dfrac{3}{7}+\dfrac{2}{67}\right)\\ =\dfrac{2}{67}-\dfrac{215}{469}\\ =\dfrac{-3}{7}\)

\(\dfrac{3}{4}.\dfrac{5}{2}+\dfrac{5}{4}.\dfrac{5}{2}\\ =\left(\dfrac{3}{4}+\dfrac{5}{4}\right).\dfrac{5}{2}\\ =2.\dfrac{5}{2}\\ =\dfrac{5}{1}\)

Ta có:

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

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

\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{x^3-y^3}\)

\(=\dfrac{2x^2+2y^2-4xy}{x^3-y^3}\)

\(=\dfrac{2x^2-2xy-2xy+2y^2}{x^3-y^3}\)

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

\(=\dfrac{\left(2x-2y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

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