a) \(\dfrac{3x^2+6xy}{6x^2}=\dfrac{3x\left(x+2y\right)}{6x^2}=\dfrac{x+2y}{2x}\)

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

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

`a, (3x^2+6xy)/(6x^2) = (x+2y)/(3x)`

`b, (2x^2-x^3)/(x^2-4) = (x^2(2-x))/((x-2)(x+2))`

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

`c, (x+1)/(x^3+1) = 1/(x^2-x+1)`

a) \(\dfrac{11\cdot8-11\cdot3}{17-6}\)

\(=\dfrac{11\cdot\left(8-3\right)}{11}=5\)

b) \(\dfrac{24-12\cdot13}{12+4\cdot9}\)

\(=\dfrac{12\cdot\left(2-13\right)}{12\left(1+3\right)}=\dfrac{-11}{4}\)

\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1}=\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{24}\left(x^3+1\right)+x^{18}\left(x^3+1\right)+x^{12}\left(x^3+1\right)+x^6\left(x^3+1\right)+\left(x^3+1\right)}\)

=\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\frac{1}{x^3+1}\)

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

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

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

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

b)\(\dfrac{x^3-36x}{x^2+6x}\)

=\(\dfrac{x\left(x^2-6^2\right)}{x\left(x+6\right)}\)

=\(\dfrac{x\left(x+6\right)\left(x-6\right)}{x\left(x+6\right)}\)

= \(x-6\)

#Fiona 

Chúc bạn học tốt !

a)      Cách 1:

\(6(y - x) - 2(x - y)\)

\( = 6y - 6x - 2x + 2y\)

\( = 8y - 8x\)

Cách 2:

\(6(y - x) - 2(x - y)\\= 6(y-x)+2(y-x)\\=(6+2).(y-x)\\=8.(y-x)\\=8y-8x\)

b)      \(3{x^2} + x - 4x - 5{x^2}\)

\( = (3{x^2} - 5{x^2}) + (x - 4x)\)

\( =  - 2{x^2} - 3x\)

a. \(\left|2x-4\right|+\left|x-3\right|\)

Với \(x< 2\), biểu thức trở thành 

\(-\left(2x-4\right)-\left(x-3\right)\)

\(=-2x+4-x+3\)

\(=-3x+7\)

Với \(2\le x< 3\), biểu thức trở thành

\(\left(2x-4\right)-\left(x-3\right)\)

\(=2x-4-x+3\)

\(=x-1\)

Với \(x\ge3\), biểu thức trở thành

\(\left(2x-4\right)+\left(x-3\right)\)

\(=2x-4+x-3\)

\(=3x-7\)

b. \(\left|x-5\right|+\left|x+6\right|\)

Với \(x< -6\), biểu thức trở thành

\(-\left(x-5\right)-\left(x+6\right)\)

\(=-x+5-x-6\)

\(=-2x-1\)

Với \(-6\le x< 5\), biểu thức trở thành

\(-\left(x-5\right)+\left(x+6\right)\)

\(=-x+5+x+6\)

\(=11\)

Với \(x\ge5\), biểu  thức trở thành

\(\left(x-5\right)+\left(x+6\right)\)

\(=x-5+x+6\)

\(=2x+1\)

a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)

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

c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)

d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)

`a, (3x^2y)/(2xy^5)`

`= (3x)/(2y^4)`

`b, (3x^2-3x)/(x-1)`

`= (3x(x-1))/(x-1)`

`= 3x`

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

`= (b(a-b))/((2a+1))`

`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.

a) 2x(x+3) – 3x²(x+2) + x(3x² + 4x – 6)

= (2x . x + 2x . 3) – (3x² . x + 3x² . 2) + (x . 3x² + x . 4x – x . 6)

= 2x² + 6x – (3x³ + 6x²) + (3x³ + 4x² - 6x)

= 2x² + 6x – 3x³ – 6x² + 3x³ + 4x² - 6x

= (– 3x³ + 3x³ ) + (2x²  - 6x² + 4x² ) + (6x – 6x)

= 0 + 0 + 0

= 0

b) 3x(2x² – x) – 2x²(3x+1) + 5(x² – 1)

= [3x . 2x² + 3x . (-x)] – (2x² . 3x + 2x² . 1) + [5x² + 5 . (-1)]

= 6x³ – 3x² – (6x³ +2x²) + 5x² – 5

= 6x³ – 3x² – 6x³ - 2x² + 5x² – 5

= (6x³ – 6x³ ) + (-3x² – 2x² + 5x²) – 5

= 0 + 0 – 5

= - 5

a: x+31+40=x+71

b: =-35-x-23-61=-119-x