giúp tui 

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

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

a: =4(x-2)(x+1)+4(x-2)^2+(x+1)^2

=(2x-4)^2+2*(2x-4)(x+1)+(x+1)^2

=(2x-4+x+1)^2=(3x-3)^2=9(x-1)^2

b: =x^7(x^2-1)-x^5(x+1)+x^3(x+1)+(x^2-1)

=(x+1)[x^7(x-1)-x^5+x^3+x-1]

=(x+1)[x^7(x-1)-x^3(x-1)(x+1)+(x-1)]

=(x+1)(x-1)(x^7-x^4-x^3+1)

=(x+1)(x-1)(x^3-1)(x^4-1)

=(x+1)(x-1)^2*(x^2+x+1)(x^2+1)(x-1)(x+1)

=(x+1)^2*(x-1)^3*(x^2+1)(x^2+x+1)

Đặt A=\(\left(x+3\right)^4+\left(x+5\right)^4-2\)

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

Đặt b=x+4

=>\(A=\left(b-1\right)^4+\left(b+1\right)^4-2\)

\(=\left(b^2-2b+1\right)^2+\left(b^2+2b+1\right)^2-2\)

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

\(=2\left(b^2+1\right)^2+8b^2-2\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-1\right\rbrack\)

\(=2\cdot\left\lbrack b^4+2b^2+1+4b^2-1\right\rbrack=2\left(b^4+6b^2\right)=2b^2\left(b^2+6\right)\)

\(=2\left(x+4\right)^2\left\lbrack\left(x+4\right)^2+6\right\rbrack\)

Đặt A=\(\left(x+3\right)^4+\left(x+5\right)^4-2\)

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

Đặt b=x+4

=>\(A=\left(b-1\right)^4+\left(b+1\right)^4-2\)

\(=\left(b^2-2b+1\right)^2+\left(b^2+2b+1\right)^2-2\)

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

\(=2\left(b^2+1\right)^2+8b^2-2\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-1\right\rbrack\)

\(=2\cdot\left\lbrack b^4+2b^2+1+4b^2-1\right\rbrack=2\left(b^4+6b^2\right)=2b^2\left(b^2+6\right)\)

\(=2\left(x+4\right)^2\left\lbrack\left(x+4\right)^2+6\right\rbrack\)

=x^3(x^2+x+1)+(x^2+x+1)

=(x^2+x+1)(x^3+1)

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

\(x^5+x^4+x^3+x^2+x+1\)

\(=\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)\)

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

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

#Toru

  (x+2).(x+3)(x+4)(x+5)-24

=(x+2)(x+5)(x+3)(x+4)-24

=(x²+7x+10)(x²+7x+12)-24

=(x²+7x+11-1)(x²+7x+11+1)-24

Đặt x²+7x+11=a thì

=(a-1)(a+1)-24

=a²-1-24=a²-25=a²-5²

=(a+5)(a-5)

=(x²+7x+16)(x²+7x+6)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25\)

\(=\left(x^2+7x+11\right)^2-25=\left(x^2+7x+11+5\right)\left(x^2+7x+11-5\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

=[(x+2)(x+5)][(x+3)(x+4)]−24=(x2+7x+10)(x2+7x+12)−24=(x2+7x+11)2−1−24=(x2+7x+11)2−25=(x2+7x+11−5)(x2+7x+11+5)=(x2+7x+6)(x2+7x+16)=(x+1)(x+6)(x2+7x+16)

( x+3) (x+2) (x+4) (x+5) -24

=(x+3)(x+4)(x+2)(x+5)-24

=(x²+7x+12)(x²+7x+10)-24

Đặt t=x²+7x+10 ta được:

(t+2)t-24

=t²+2t-24

=t²+4t-6t-24

=t.(t+4)-6.(t+4)

=(t+4)(t-6)

thay t=x²+7t+10 ta được:

(x²+7x+14)(x²+7+4)

Vậy  ( x+3) (x+2) (x+4) (x+5) -24=(x²+7x+14)(x²+7x+4)

song la phai nhuong nhin

ban ghep (x+3)(x+4) thanh 1 cap va (x+2)(x+5) roi dat t=x^2+7x+10                                                                    ta co t(t+2)-24=(t^2 +2t+1) -25=(t+1)^2 - 5^2=(t+6)(t-4)                                                                                       thay t la xong . giup mk lam bai mk dang vs nhe!