a) 2x - x³ + 4y - 8y³

= ( 2x + 4y ) - ( x³ + 8y³ )

= 2( x + 2y ) - ( x + 2y )( x² - 2xy + 4y² )

= ( x + 2y )( 2 - x² + 2xy - 4y² )

b) -3x² + 11x + 14

= -3x² - 3x + 14x + 14

= -3x( x + 1 ) + 14( x + 1 )

= ( x + 1 )( 14 - 3x )

a) 2x - x³ + 4y - 8y³ 

= (2x + 4y) - (x³ + 8y³) 

= 2 (x + y) - [x³ + (2y)³] 

= 2 (x + y) - (x + y)(x² - 2xy + 4y²) 

= (x + y)( 2 - x² + 2xy - 4y²)      (Thật sự là câu này mình vẫn chưa chắc chắn lắm =)))

b) -3x² + 11x + 14 

= -3x² - 3x + 14x + 14 

= (-3x² - 3x) + (14x + 14) 

= -3x(x + 1) + 14(x + 1) 

= (-3x + 14)(x + 1)

=))

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

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

\(=\left(x-2y-3\right)\left(x+2y\right)\)

b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)

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

c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)

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

d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)

a) \(\dfrac{4y^2}{11x^4}:\left(-\dfrac{8y}{33x^2}\right)\)

\(=\dfrac{4y^2}{11x^4}.\left(-\dfrac{33x^2}{8y}\right)\)

\(=-\dfrac{4y^2.33x^2}{11x^4.8y}\)

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

b) \(\dfrac{x^2-4}{3x+12}.\dfrac{x+4}{2x-4}\)

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

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

\(=\dfrac{x+2}{6}\).

\(\dfrac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}=\dfrac{x+2}{6}\)

\(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}\)

\(=\frac{5}{4}.\frac{-2}{1}=\frac{-10}{4}\)

a) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}=\dfrac{15x.2y^2}{7y^3.x^2}=\dfrac{30}{7xy}\)

b) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)=\dfrac{-4y^2.3x^2}{11x^4.8y}=\dfrac{-3y}{22x^2}\)

c) \(\dfrac{x^3-8}{5x+20}.\dfrac{x^2+4x}{x^2+2x+4}\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}.\dfrac{x\left(x+4\right)}{x^2+2x+4}\\ =\dfrac{x^2-2x}{5}\)

A) 3x² - x(3x - 5) = 9

3x² - 3x² + 5x = 9

5x = 9

x = 9/5

--------------------

B) 5x² + 9x - 2 = 0

5x² + 10x - x - 2 = 0

(5x² + 10x) - (x + 2) = 0

5x(x + 2) - (x + 2) = 0

(x + 2)(5x - 1) = 0

x + 2 = 0 hoặc 5x - 1 = 0

*) x + 2 = 0

x = -2

*) 5x - 1 = 0

5x = 1

x = 1/5

Vậy x = -2; x = 1/5

---------------------

D) 4(5 - 3x) = 5x - 5

20 - 12x = 5x - 5

-12x - 5x = -5 - 20

-17x = -25

x = 25/17

--------------------

E) 2x² - 11x + 14 = 0

2x² - 4x - 7x + 14 = 0

(2x² - 4x) - (7x - 14) = 0

2x(x - 2) - 7(x - 2) = 0

(x - 2)(2x - 7) = 0

x - 2 = 0 hoặc 2x - 7 = 0

*) x - 2 = 0

x = 2

*) 2x - 7 = 0

2x = 7

x = 7/2

Vậy x = 2; x = 7/2

Câu C và F ghi đề bằng công thức đúng lại em

c: =>x^2-(x-2)(x+5)=2x-1

=>x^2-x^2-5x+2x+10=2x-1

=>3x+10=2x-1

=>x=-11

f: =>3x-5=4(2x+3)

=>8x+12=3x-5

=>5x=-17

=>x=-17/5

\(a,\frac{4y^2}{11x^4}.\frac{3x^2}{3y}=\frac{4y}{11x^2}\)

\(b,\left(x^2-49\right):\frac{2x+14}{3x-5}=\frac{1}{\left(x-7\right)\left(x+7\right)}.\frac{2\left(x+7\right)}{3x-5}\)

\(=\frac{2}{\left(x-7\right)\left(3x-5\right)}\)

1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

a) Ta có: \(x-2y+x^2-4y^2\)

\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)

\(=\left(x-2y\right)\left(x+2y+1\right)\)

b) Ta có: \(x^2+2xy+y^2-4x^2y^2\)

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

\(=\left(x+y+2xy\right)\left(x+y-2xy\right)\)

c) Ta có: \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

d) Ta có: \(x^3+3x^2+3x+1-8y^3\)

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

\(=\left(x+1-2y\right)\left[\left(x+1\right)^2+2y\left(x+1\right)+4y^2\right]\)

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