Chứng minh rằng: `a^2+ab+b^2=3/4(a+b)^2+1/4(a-b)^2` `a^2-ab+b^2=3/4(a-b)^2+1/4(a+b)^2` `(a+b)^3-(a-b)^3=ab(b^2+3a^2)`

Đáp án:+Giải thích các bước giải:

 Ta có:

`VP=3/4 (a+b)^2 + 1/4  (a-b)^2`

`    =3/4 (a^2 +2ab+b^2)+1/4 (a^2 -2ab+b^2)`

`    =3/4 a^2 +3/2 ab + 3/4 b^2 +1/4 a^2 - 1/2 ab+1/4 b^2`

`    =(3/4 a^2 +1/4 a^2)+(3/2 ab - 1/2 ab)+(3/4 b^2 +1/4 b^2)`

`    =a^2 +ab +b^2  =VT`

Vậy `a^2 +ab+b^2 =3/4 (a+b)^2 + 1/4 (a-b)^2`

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Ta có: 

`VP=3/4 (a-b)^2 + 1/4 (a+b)^2`

`    =3/4 (a^2 -2ab+b^2)+1/4 (a^2 +2ab+b^2)`

 `   =3/4 a^2 - 3/2 ab +3/4 b^2 +1/4 a^2 +1/2 ab + 1/4 b^2`

`    =(3/4 a^2 + 1/4 a^2)+(-3/2 ab + 1/2 ab)+(3/4 b^2 + 1/4 b^2)`

 `   =a^2 -ab + b^2 =VT`

Vậy `a^2 -ab+b^2 =3/4 (a-b)^2 + 1/2 (a+b)^2`

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Ta có:

Sửa `VP=2b(b^2 +3a^2)`

`VT=(a+b)^3 -(a-b)^3`

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

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

`     =6a^2 b +2b^3`

   `  =2b(b^2+3a^2) =VP`

Vậy `(a+b)^3 -(a-b)^3 =ab(b^2 +3a^2)`