1) \(\left(a+b\right)^2\)

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

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

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

2) \(\left(a-b\right)^3\)

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

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

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

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

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

`a)` 

`(a+b)^2`

`=(a+b)(a+b)`

`=a^2+ab+ab+b^2`

`=a^2+2ab+b^2`

`->` ĐPCM

`b)` `(a-b)^3`

`=(a-b)(a-b)(a-b)`

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

`=a^3-3a^2b+3ab^2-b^3`

`->` ĐPCM

Biến đổi VP

=> VT = VP

=> Đpcm

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

Không có được CM ngược lại bạn ạ

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

Ta xét vế phải trước :

 VP = ( a + b ) ³ - 3ab ( a + b )

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

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

\(=a^3+b^3=VT\)( đpcm )

1) biến đổi vế trái:

= a²+2ab+b² -a² +2ab -b²

=4ab = vế phải ( đpcm)

3;5 tuong tu

1) (a + b)² - (a - b)² = a² + 2ab + b² - a² + 2ab - b² = 4ab

3) (a + b)² - 4ab = a² + 2ab + b² - 4ab = a² - 2ab + b² = (a - b)²

5) a³ + b³ = a³ + 3a²b + 3ab² + b³ - 3a²b - 3ab² = (a + b)³ - 3ab(a + b)

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

VT= (a + b - a +b) (a+b + a-b) 

= 2b * 2a 

= 4ab = VP

Vậy (a+b)^2 - (a-b)^2 = 4ab

2) (a+b)^2 - 4ab = (a-b)^2 

VT= (a+b)^2 - 4ab 

= a^2 + 2ab + b^2 - 4ab

= a^2 - 2ab + b^2 

= (a-b)^2  = VP

Vậy (a+b)^2 - 4ab = (a-b)^2 

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

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

\(=a^3+b^3\)

Lời giải:

$(a+b)^3-3ab(a+b)$

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

$=a^3+b^3$
Ta có đpcm.

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

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

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

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

\(=a^3+b^3\)