Question

chứng minh đẳng thức sau

a) (a-b)^3=  -(b-a)^3

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

chứng minh hộ mình vs ạ

 

Answer 1

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

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

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

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

`=> (a+b)^2 = (a+b)^2`.

Answer 2

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

Vậy \(\left(a-b\right)^3=-\left(b-a\right)^3\)

b) VT = \(\left(-a-b\right)^2=\left[\left(-1\right)\left(a+b\right)\right]^2=\left(-1\right)^2\left(a+b\right)^2=\left(a+b\right)^2\) = VP

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

 

Answer 3

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

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