\(x^6-y^6\)

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

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

\(x^6-y^6\)

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

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

a)\(A=1^3+2^3+3^3+........+10^3\)

\(A=1^3+10^3+2^3+9^3+3^3+8^3+4^3+7^3+5^3+6^3\)

\(A=11\cdot111+11\cdot103+11\cdot97+11\cdot93+11\cdot91\)

\(A=11\cdot\left(111+103+97+93+91\right)=11\cdot495\)

\(A=11\cdot11\cdot5\cdot9\)

Vậy \(A⋮11,A⋮5\)

Mình chưa hiểu difng 3 cho lắm. Tại sao lại có 11.111 vậy? 

\(a^3+3a^2b+3ab^2+b^3-8c^3\)

\(=\left(a+b\right)^3-8c^3\)

\(=\left(a+b-2c\right)\left[\left(a+b\right)^2+2\left(a+b\right)c+4c^2\right]\)

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

thank bạn

\(x^2-16+\left(x-3\right)^2\)

\(=x^2-16+x^2-6x+9\)

\(=2x^2-6x-7\)

___Cạn , xem lại đề coi_

_Minh ngụy_

#)Giải :

Ta có :

\(mn\left(m^2-n^2\right)=mn\left[\left(m^2-1\right)-\left(n^2-1\right)\right]=n\left\{m\left[m^2-1\right]-m\left[n\left(n^2-1\right)\right]\right\}\)

\(=mn\left(m-1\right)\left(m+1\right)-mn\left(n-1\right)\left(n+1\right)\)

\(m\left(m-1\right)\left(m+1\right)⋮6\left(1\right)\)

\(n\left(n-1\right)\left(n+1\right)⋮6\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow mn\left(m-1\right)\left(m+1\right)-mn\left(n-1\right)\left(n+1\right)⋮6\)

\(\Rightarrow mn\left(m^2-n^2\right)⋮6\)

Mà \(4mn\left(m^2-n^2\right)⋮4\)

\(\Rightarrow4mn\left(m^2-n^2\right)⋮24\left(đpcm\right)\)

Trả lời

2002 x 1006

= ( 1504 + 498 ) x ( 1504 - 498 )

= 1504² - 498²

= 2014012

198 x 202 

= ( 200 - 2 ) x ( 200 + 2 )

= 202² - 2²

= 40800

#)Giải :

\(\left(-2a-4\right)\left(2a+1\right)\)

\(=\left(-2a\right)\left(2a+1\right)+4\left(2a+1\right)\)

\(=\left(-2a\right).2a+\left(-2a\right).1+4.2a+4.1\)

\(=-4a^2-2a+8a+4\)

\(=-4a^2+6a+4\)

\(a)-64+y^2=\left(y-8\right)\left(y+8\right)\)

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

- 64 + y ²

= - 8 . 8 + y . y

= ( y + 8 ) . ( y - 8 ) 

-27 - x³

= - ( 27 + x³ )

= - ( 3 + x ) . ( 9 - 3 + x² )

a) \(x^3-16x=0\)

\(\Leftrightarrow x\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{16}=\pm4\end{cases}}\)

Vậy \(x\in\left\{0;\pm4\right\}\)

b) \(x^2-6x+9=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)