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

b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

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

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

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

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

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

ta có : 

\(a^3+c^3=\left(a+c\right)^3-3ac\left(a+c\right)\)

nên \(a^3+c^3-b^3+3abc=\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)\)

\(=\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2-3ac\right]=\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)\)

b. tương tự ta có :

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

\(=\left(a-b-c\right)\left[a^2+a\left(b+c\right)+\left(b+c\right)^2-3bc\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)

c. ta có : \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=\left(x-z+z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\(=\left(x-z\right)^3+3\left(x-z\right)\left(z-y\right)\left(x-y\right)+\left(z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\(=3\left(x-z\right)\left(z-y\right)\left(x-y\right)\)

1) 

a) (x+y)³-(x+y)= (x+y)(x+y-1)

b) xem lại đề câu B nha bạn

2)

a³+3a²b+3ab²+b³+c³-3a²b-3ab²-3abc=0

(a+b)³+c³-3ab(a+b+c)=0

(a+b+c)(a²+2ab+b²-ac-bc+c²)-3ab(a+b+c)=0

(a+b+c)(a²+b²+c²-xy-yz-xz)=0

Suy ra: a³+b³+c³=3abc

1. a) = (x+y)³ -(x+y) =(x+y)((x+y)² -1)

     = (x+y)(x+y+1)(x+y-1)

b) = 5(( x-y)² - 4z²)

     = 5( x-y +2z)(x-y-2z)

2. áp dụng ( a+b+c)³ = .....rồi biến đổi

 ta có a³+b³+c³-3abc

        =a³+3a²b+3ab²+b³+c³-3a²b-3ab²-3abc

        =(a+b)³+c³-3ab(a+b+c)

        =(a+b+c)((a+b)²-c(a+b)+c²-3ab)

        =0((a+b)²-c(a+b)+c²)

        =0

suy ra a³+b³+c³ -3abc=0

          a³+b³+c³=3abc(đpcm)

a)x(x+1)\(^2\)

b)(y-1)(x+y)

Ta có : x³ + 2x² + x 

= x³ + x² + x² + x

= x²(x + 1) + x(x + 1)

= (x² + x) (x + 1)

= x(x + 1)(x + 1)

Ta có : xy + y² - x - y 

= y(x + y) - (x + y)

= (x + y)(y - 1)

Bài 1 :

a) xy(x+y)+yz(y+z)+xz(x+z)+2xyz 

= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

= xy(x + y) + yz(y + z + x) + xz(x + z + y) 

= xy(x + y) + z(x + y + z)(y + x) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

= (x + y)(y + z)(z + x)

b) \(x^3-x+3x^2y+3xy^2+y^3-x-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)

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

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

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

Đã có kết quả

Bài 1,chữa phần a

 xy(x+y)+yz(y+z)+xz(x+z)+2xyz

=[xy(x+y)+xyz]+[yz(y+z)+xyz]+xz(x+z)

=xy(x+y+z)+yz(x+y+z)+xz(x+z)

=y(x+y+z)(x+z)+xz(x+z)

=(x+z)(xy+y²+yz+xz)

=(x+z)(x+y)(y+z)

Chữa phần b

x³-x+3x²y+3xy²+y³-y

=(x+y)(x+y-1)(x+y+1)

Bài2

a³+b³+c³=(a+b)³-3ab(a+b)+c³=-c³-3ab(-c)+c³=3abc

Ai làm đúng như này ớ sẽ k

Mình làm bài 2 luôn:

  Ta có: a³+b³+c³=(a+b)³-3a²b-3ab²+c³

                          =(a+b)³+c³-3ab(a+b)

                          =(a+b+c)[(a+b)²-(a+b)c+c² ]-3ab(a+b)

                          =-3ab(a+b)    (vì a+b+c=0)

Từ a+b+c=0 =>-c=a+b

  => -3ab(a+b)=-3ab(-c)=3abc     

                                                  đpcm

a) \(x^2+2x+1=x^2+x+x+1=x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x+1\right)=\left(x+1\right)^2\)    *Câu này có thể áp dụng hằng đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\)  cho nhanh*

b) \(a^3-b^3+c^3+3abc=\left(a^3-3a^2b+3ab^2-b^2\right)+3a^2b-3ab^2+c^3+3abc\)

\(=\left(a-b\right)^3+c^3+\left(3a^2b-3ab^2+3abc\right)\) 

\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)

\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)

\(=\left(a-b+c\right)\left(a^2+b^2+c^2-ac+bc+ab\right)\)

c) \(a^3-b^3-c^3-3abc=\left[a^3-3a^2b+3ab^2-b^3\right]+3a^2b-3ab^2-c^3-3abc\)

\(=\left[\left(a-b\right)^3-c^3\right]+3ab\left(a-b-c\right)=\left(a-b-c\right)\left[\left(a-b\right)^2+\left(a-b\right)c+c^2\right]+3ab\left(a-b-c\right)\)

\(=\left(a-b-c\right)\left[a^2-2ab+b^2+ac-bc+c^2+3ab\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)

a,(x+1)²

b,(a+c-b).{(a+c)^2+(a+c)b+b^2-3ac}

c,(a-c-b).{(a-c)^2+(a-c)b+b^2+3ac}