Đặt n^2 + 2n + 1= a, ta được:

(a - 1)(a + 1) +1= a^2  -  1 + 1= a^2=(n^2 + 2n +1)^2

=(n + 1)^4

a: \(M=m^2\left(m+n\right)-n^2m-n^3\)

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

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

\(=\left(-2017+2017\right)^2\cdot\left(-2017-2017\right)\)

=0

b: \(N=n^3-3n^2-n\left(3-n\right)\)

\(=n^2\left(n-3\right)+n\left(n-3\right)\)

\(=n\left(n-3\right)\left(n+1\right)\)

\(=13\cdot10\cdot14=1820\)

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)

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

Viết liền thế ai mà đọc được

Khó quá

B1:

[(m+n)+(2m-3n)]^2

= (m+n)^2 + 2(m+n)(2m-3n) + (2m-3n)^2

= m^2 +2mn +n^2 + 4m^2 - 6mn + 4mn - 6n^2 + 4m^2 - 12mn + 9n^2

= 9m^2 - 12mn + 4n^2
 

B2,3

 bn lm theo hdt ( a +b + c) ^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc nha

a)\(m+n+\left(2m-3n\right)^2=\left(3m-2n\right)^2\)

b)(x+y+z)^2 đã thành nhân tử 

c)(2x-y+3z)^2 đã thành nhân tử, too

d)(x+y)(x-y)(2x+y).........

e)\(8x^6-y^6=\left(2x^2-y^2\right)\left(4x^4+2x^2y^2+y^4\right)\)

g)\(\left(m+n\right)^3-\left(2m+3n\right)^3\)

\(=-\left(m+2n\right)\left(7m^2+19mn+13n^2\right)\)

a: Ta có: \(m^2+2mn+n^2-p^2+2pq+q^2\)

\(=\left(m+n\right)^2-\left(p-q\right)^2\)

\(=\left(m+n-p+q\right)\left(m+n+p-q\right)\)