Question

[Cỏ dại]

Tính giá trị biểu thức:

\(x^4+4x^3y+6x^2y^2+4xy^3+y^4-x-y-10\) với x + y = 2

Answer 1

[Không Tên]

         \(x^4+4x^3y+6x^2y^2+4xy^3+y^4-x-y-10\)

\(=\left(x^4+2x^3y+x^2y^2\right)+\left(2x^3y+4x^2y^2+2xy^3\right)+\left(x^2y^2+2xy^3+y^4\right)-\left(x+y\right)-10\)

\(=x^2\left(x^2+2xy+y^2\right)+2xy\left(x^2+2xy+y^2\right)+y^2\left(x^2+2xy+y^2\right)-\left(x+y\right)-10\)

\(=\left(x^2+2xy+y^2\right)\left(x^2+2xy+y^2\right)-\left(x+y\right)-10\)

\(=\left(x+y\right)^2\left(x+y\right)^2-\left(x+y\right)-10\)

\(=\left(x+y\right)^4-\left(x+y\right)-10\)

\(=2^4-2-10\) \(=4\)