Question

[Tran Thi Kim Phung]

Tính gtbt:

a)

x^3+9x^2y+27xy^2+27y^3

Tại 1/3x+y+1=0

b)

[3^(x+y)^2]/[3^(x-y)^2

tại xy=1/2

c)

2(x^3-y^3)-3(x+y)^2

tại x-y=2

Answer 1

[Devil]

a)

ta có: 1/3x+y+1=0

=> 1/3x+y=-1

ta có : x^3+9x^2y+27xy^2+27y^3

=(x+3y)^3

= [1/3(1/3x+y)]^3

=1/27 . (-1)^3

=1/27 . (-1)

=-1/27

Answer 2

[Devil]

\(b\frac{3^{x+y^2}}{3^{x-y^2}}=\frac{3^{x^2+2xy+y^2}}{3^{x^2-2xy+y^2}}=\frac{3^{x^2+y^2}.3^{2xy}}{3^{x^2+y^2}:3^{2xy}}=3^{2xy}.3^{2xy}=3^{\frac{4.1}{2}}=3^2=9\)