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

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

Khó quá

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: \(N=n^3-3n^2-n\left(3-n\right)\)

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

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

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

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

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

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

\(2m^2+10m+8\)

\(=2\left(m^2+5m+4\right)\)

\(=2\left(m^2+4m+m+4\right)\)

\(=2\left(m+4\right)\left(m+1\right)\)

=2m²+8m+2m+8

=(2m²+2m)+(8m+8)

=2m(m+1)+8(m+1)

=(m+1)(2m+8)

=(m+1)2(m+4)

=2(m+1)(m+4)

HT~

\(3m^2-2m-1\)

\(=3m^2-3m+m-1\)

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

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

\(3m^2-2m-1\)

\(=3m^2+m-3m-1\)

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

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

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

m^2(n-p) + n^2(p-m) + p^2(m-n)

= m²n-m²p +n²p-n²m+p²(m-n)

= mn(m-n) -p(m²-n²)+p²(m-n)

= mn(m-n) -p(m-n)(m+n)+p²(m-n)

=(m-n)(mn-pm-pn+p²)

=(m-n)[m(n-p)-p(n-p)]

=(m-n)(m-p)(n-p)