2. Chứng minh rằng:
a. a mũ 3 + b mũ 3 = (a + b) mũ 3 – 3ab (a + b)
b. a mũ 3 + b mũ 3 + c mũ 3 – 3abc = (a + b + c) (a mũ 2 + b mũ 2 c mũ 2 – ab – bc – ca)
2. Chứng minh rằng: a. a mũ 3 + b mũ 3 = (a + b) mũ 3 – 3ab (a + b) b. a mũ 3 + b mũ 3 + c mũ 3 – 3abc = (a + b + c) (a mũ 2 + b mũ 2 c mũ 2 – ab – bc
By Kylie
Đáp án:
`a)`
Ta có
` (a+b)^3 – 3ab(a+b)`
` = a^3 + 3a^2b+3ab^2 + b^3 – 3a^2b – 3ab^2`
` = a^3 +b^3 + (3a^2b – 3a^2b) + (3ab^2 – 3ab^2)`
` = a^3 +b^3 + 0 +0`
` = a^3 +b^3`
`b)`
` (a+b+c)(a^2+b^2+c^2 -ab – bc -ac)`
` = a^3 + ab^2 +ac^2 – a^2b -abc – a^2c + a^2b + b^3 + bc^2 – ab^2 – b^2c – abc + a^2c + b^2c + c^3 – abc – bc^2 – ac^2`
` = (a^3 + b^3 + c^3 – 3abc) + (ab^2 – ab^2) – (a^2b-a^2b) – (a^2c-a^c) + (b^2c – b^2c) + (bc^2 – bc^2)`
` = a^3 + b^3 + c^3 -3abc`
a) Ta có
$VT = (a+b)^3 – 3ab(a+b) = a^3 + b^3 + 3a^2b + 3ab^2 – 3a^2b – 3ab^2$
$= a^3 + b^3 = VP$.
b) Ta có
$VT = a^3 + b^3 + c^3 – 3abc$
$= (a+b)^3 – 3ab(a+b) + c^3 – 3abc$
$= [(a+b)^3 + c^3] – 3ab(a + b + c)$
$= (a+b+c)[(a+b)^2 + c^2 – c(a+b)] -3ab(a+b+c)$
$= (a+b+c)(a^2 + b^2 + 2ab + c^2 – ac – bc – 3ab)$
$= (a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca) = VP$