Đề sai rồi bạn

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)

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

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

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

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}

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

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

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

a: \(x^2-y^2+10yz-25z^2\)

\(=x^2-\left(y^2-10yz+25z^2\right)\)

\(=x^2-\left(y-5z\right)^2\)

\(=\left(x-y+5z\right)\left(x+y-5z\right)\)

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

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

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

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

bài a) bn trên đã dẫn link cho bn r

bài b)

Đặt x-y=a;y-z=b;z-x=c 

\(=>a+b+c=x-y+y-z+z-x=0\)

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

Theo câu a)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) (do a+b+c=0)

\(=>a^3+b^3+c^3=3abc=>\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

a) Ta có :

\(a^3+b^3+c^3-3abc\)

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

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

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

P/s tham khảo nha

hok tốt