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

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

c)

\(\frac{9x^2y^2+3x^2}{12xy^5+4xy^3}=\frac{3x^2\left(3y^2+1\right)}{4xy^3\left(3y^2+1\right)}=\frac{3x^2}{4xy^3}\)

@@@@ỦNG HỘ NHOA@@@@

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

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

= x(x - 2y - 3)(x - 2y + 3)

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

= (3x - 2y)(3x + 2y) - (3x - 2y)

= (3x - 2y)(3x + 2y - 1)

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

https://www.youtube.com/watch?v=cFZDEMTQQCs

Học tốt <3

x(3x-1)+(9x-5)(x-2)=3x²-x+9x(x-2)-5(x-2)=3x²-x+9x²-18x-5x+10=12x²-22x+10

a)4x²+12xy+9y²=  (2x)²+2.2x.3y+(3y)²=(2x+3y)²   

b)y²+1-2y= y²-2.y.1+1²=(y-1)²

\(=\dfrac{3\left(x+1\right)\left(3x-5\right)}{-\left(3x-5\right)\left(3x+5\right)}=\dfrac{-3\left(x+1\right)}{3x+5}\)

\(a.\)

\(\dfrac{16x^2-1}{16x^2-8x+1}\\ =\dfrac{\left(4x\right)^2-1}{\left(4x-1\right)^2}\\ =\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\\ =\dfrac{4x+1}{4x-1}\)

\(b.\)

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

a) Ta có: \(\dfrac{16x^2-1}{16x^2-8x+1}\)

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

\(=\dfrac{4x+1}{4x-1}\)

b) Ta có: \(\dfrac{4x^2-4xy+y^2}{y^2-4x^2}\)

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

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

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

a: \(=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2\)

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

d: \(=9x^2+6x+1-9x^2+6x-1=12x\)