a) \( - 2{x^2} + 6{x^2} = ( - 2 + 6).{x^2} = 4{x^2}\);                  

b) \(4{x^3} - 8{x^3} = (4 - 8).{x^3} =  - 4{x^3}\);

c) \(3{x^4}( - 6{x^2}) = 3.( - 6).{x^4}.{x^2} =  - 18{x^{4 + 2}} =  - 18{x^6}\);                 

d) \(( - 24{x^6}):( - 4{x^3}) = ( - 24: - 4).({x^6}:{x^3}) = 6{x^{6 - 3}} = 6{x^3}\). 

`x + 2 3/4 = 5 2/3`

`=> x + 11/4 = 17/3`

`=> x= 17/3 -11/4`

`=>x=35/12`

__

`x - 1 4/5 = 3 2/7`

`=> x- 9/5 = 23/7`

`=> x= 23/7 +9/5`

`=>x=178/35`

__

`x xx 3 1/2 =4 3/4`

`=> x xx 7/2 =19/4`

`=> x= 19/4 : 7/2`

`=> x= 19/4 xx 2/7`

`=> x= 19/14`

__

`x : 2 2/3 = 4 1/3`

`=> x : 8/3 = 13/3`

`=> x= 13/3 xx 8/3`

`=>x=104/9`

a: =>x+11/4=17/3

=>x=17/3-11/4=68/12-33/12=35/12

b: =>x-9/5=23/7

=>x=23/7+9/5=178/35

c: =>x*7/2=4,75

=>x=19/4:7/2=19/14

d: =>x:8/3=13/3

=>x=13/3*8/3=104/9

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

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

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

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

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

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

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

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

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)

$=x^3-2x^5$

b) $(x^2+1)(5-x)$

$=5x^2-x^3+5-x$

$=-x^3+5x^2-x+5$

$=x^3+3x^2-4x-2x^2-6x+8$

$=x^3+x^2-10x+8$

$=x^2-x^3+4x-2x+2x^2-8$

$=-x^3+3x^2+2x-8$

$=x^4+2x^3-x^2-2x$

Trước hết:

$(3x+2)(3-x)=9x+6-3x^2-2x$

$=-3x^2+7x+6$

Do đó:

$(2x-1)(-3x^2+7x+6)$

$=-6x^3+14x^2+12x+3x^2-7x-6$

$=-6x^3+17x^2+5x-6$

g) $(x+3)(x^2+3x-5)$

$=x^3+3x^2-5x+3x^2+9x-15$

$=x^3+6x^2+4x-15$

$=x^4y-2x^2y-6xy-2x^3+4x+12$

$=20x^5-5x^4+10x^3-4x^4+x^3-2x^2+8x^3-2x^2+4x-12x^2+3x-6$

$=20x^5-9x^4+19x^3-16x^2+7x-6$

a,`9/5+2/5xx4/6=9/5+4/15=27/15+4/15=31/15`

b,`3/8xx2-6/7xx1/3=3/4-2/7=21/18-8/28=13/28`

a) 9/5+ 4/15

= 27/15+4/15

= 31/15

b) 3/4- 2/7

= 13/28

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

b: \(=\dfrac{x-5}{\left(x-2\right)^2}\cdot\dfrac{2\left(x-2\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{\left(x-2\right)\left(x+5\right)}\)

\(a,\left(x-2\right)\left(x+3\right)-x\left(x-5\right)=x^2-2x+3x-6-x^2+5x=6x-6\)

\(b,\dfrac{1}{x-2}+\dfrac{-2}{x+2}+\dfrac{2x-8}{x^2-4}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2-2x+4+2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}\)