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

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

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

\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)

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

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

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

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

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

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

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

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

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

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

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

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

\(x^3-4x^2+12x-27\)

\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)

\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)

\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)

\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)

\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)

a, \(\left(5n+2\right)^2-4=\left(5n+2-2\right)\left(5n+2+2\right)=5n\left(5n+4\right)⋮5\)

b, \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)

Vì (n-1)n(n+1) là tích 3 số nguyên liên tiếp

=>(n-1)n(n+1) chia hết cho 6 hay n^3-n chia hết cho 6

c, \(a+b+c=0\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Rightarrow a^3+b^3-3abc=-c^3\)

=>a^3+b^3+c^3=3abc

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

                                         \(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

                                         \(=\left(x-2y-2\right)\left(x+2y\right)\)

\(x^2-2x-4y^2-4y\)

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

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

\(=\left(x+1+2y+1\right)\left(x+1-2y-1\right)\)

\(=\left(x+2y+2\right)\left(x-2y\right)\)

Bạn xem lại đề bài đi xem có chính xác k

Góc C theo đề ra có rồi .

\(\widehat{C}=120^o\)

Vậy mình  tính góc B và góc D nhá .

Vì \(AB//CD\)(giả thiết )

Ta có : \(\widehat{A}+\widehat{B}+\widehat{C} +\widehat{D}=360^o\)

Mặt khác : \(\widehat{B} +\widehat{C}=180^o\) ( đồng vị )

\(\Rightarrow\widehat{B}+120^o=180^o\)

\(\Rightarrow\widehat{B}=60^o\)

Ta lại có : \(\widehat{A}+\widehat{D}=180^o\)(kề bù )

\(\Rightarrow50^o+\widehat{D}=180^o\)

\(\Rightarrow\widehat{D}=130^o\)

Vậy góc B =60 ; góc D =130

\(-x^2+4x-2\)

\(=-x^2+4x-4+2\)

\(=-\left(x-2\right)^2+2\le2\)

Dấu "=" xảy ra khi x = 2

Vậy Max = 2 <=> x = 2