Ta có :

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

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

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

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

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

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

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

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

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

Vậy ...

trả lời lại đi bạn ơi

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

 ( 3x+2). (3x-2)+(x-3)²-10x    

=9x²-4+x²-6x+9-10x

=9x²-4+x²-6x+9

=10x-16x+5

(2x+y)²+ (x-2y)²-5. (x+y).(x-y)

=4x²+4xy+y²+x²-4xy+4y²-5.(x²-y²)

=4x²+4xy+y²+x²-4xy+4y²-5x²+5y²

=10y²

(3x-5)²- x.(3x-5)

=9x²-30x+25-3x²+15

=6x²-30x+40

mjk làm ruj đó đúng mjk đi

1/ \(P=\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2=5^2=25\)

2/\(M=a^3+b^3+3ab=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab=a^2-ab+b^2+3ab=a^2+2ab+b^2=\left(a+b\right)^2=1\)

3/

\(\left(2n+1\right)^2-\left(2n-1\right)^2=\left(2n+1-2n+1\right)\left(2n+1+2n-1\right)=2.4n=8n⋮8\)

anh jaki kìa các fan của jaki!!!

a jaki:D

b. Câu hỏi của gorosuke - Toán lớp 8 - Học toán với OnlineMath

\(a-b=5\)=> \(a=5+b\)

thay vào biểu thức P ta có

\(\left(5+b+b\right)^2-4.\left(5+b\right).b\) 

=\(\left(5+2b\right)^2-\left(20+4b\right).b\) 

= \(25+20b+4b^2-20b-4b^2\)

\(=25\)

ta có \(a+b=1\)

=> \(\left(a+b\right)^3=1\)

<=> \(a^3+3a^2b+3ab^2+b^3=1\)

<=> \(a^3+b^3+3ab.\left(a+b\right)=1\)

mà \(a+b=1\)

<=> \(a^3+b^3+3ab=1\)

hay M =1

\(\left(2n+1\right)^2-\left(2n-1\right)^2\) 

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

\(=4n^2+4n+4-\) \(4n^2+4n-1\)

\(=8n+3\)

câu cuối mk làm được thế thôi 

sorry nha

Bài 1: \(P=\left(a+b\right)^2-4ab\)

\(=a^2+2ab+b^2-4ab\)

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

\(=a^2-2ab+b^2\)

\(=\left(a-b\right)^2\)

\(=5^2\)

\(=25\)

Bài 2: \(M=a^3+b^3+3ab\)

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

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

\(=1.\left(a^2-ab+b^2\right)+3ab\)

\(=a^2-ab+b^2+3ab\)

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

\(=a^2+2ab+b^2\)

\(=\left(a+b\right)^2\)

\(=1^2=1\)

Bài 3 : Ta có : \(\left(2n+1\right)^2-\left(2n-1\right)^2\)

\(=\left(2n\right)^2+2.2n.1+1^2-\left(2n\right)^2+2.2n.1-1^2\)

\(=4.n+4.n\)

\(=8n\)Chia hết cho 8