\(\frac{2m^2+3m+1}{2m^2-m-1}=\frac{2m^2+2m+m+1}{2m^2-2m+m-1}\)

\(=\frac{2m\left(m+1\right)+\left(m+1\right)}{2m\left(m-1\right)+\left(m-1\right)}=\frac{\left(m+1\right)\left(2m+1\right)}{\left(m-1\right)\left(2m+1\right)}\)

\(=\frac{m+1}{m-1}\)

a^3m+2a^2m+a^m = a^m(a^2m+2a^m+1) = a^m[(a^m)²+2a^m+1] = a^m(a^m+1)²

Ta có :

a^3m+2a^2m+a^m 

= a^m(a^2m+2a^m+1)

= a^m[(a^m)²+2a^m+1]

= a^m(a^m+1)²

giúp me

\(=\left(m-n\right)\left(m+n\right)-2\left(m+n\right)=\left(m+n\right)\left(m-n-2\right)\)

5m^2 - 5mn + 3m - 3n

= 5m . ( m - n ) + 3 . ( m - n )

= ( m - n ) . ( 5m + 3 )

5m^2 - 5mn + 3m - 3n

= 5m(m-n) + 3(m-n)

= (m-n)(5m+3)

a) 9x² - 25

= (3x)² - 5²

= (3x - 5)(3x + 5)

b) y³ + 6y² + 9y

= y(y² + 6y + 9)

= y(y² + 2.y.3 + 3²)

= y(y + 3)²

c) m² - n² - 2m + 1

= (m² - 2m + 1) - n²

= (m - 1)² - n

= (m - 1 - n)(m - 1 + n)

= (m - n - 1)(m + n - 1)

Bài 2:

1)  \(x^2-4x+4=\left(x-2\right)^2\)

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

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

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

5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)

6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

mình cảm ơn nha