Question

Tính giá trị biểu thức biết x+y=2

\(A=x^2+y^2-x^2y-xy^2+2xy-5\)

Answer 1

\(A=x^2+y^2-xy^2-x^2y+2xy-5\)

\(=\left(x+y\right)^2-xy\left(y+x\right)-5\)

\(=2^2-2xy-5=-\left(2xy+1\right)\)

Answer 2

Trả lời:

\(A=x^2+y^2-x^2y-xy^2+2xy-5\)

\(A=\left(x^2+2xy+y^2\right)-xy.\left(x+y\right)-5\)

\(A=\left(x+y\right)^2-xy.\left(x+y\right)-5\)

\(A=2^2-xy.2-5\)

\(A=4-2xy-5\)

\(A=-1-2xy\)

\(A=-\left(1+2xy\right)\)

Học tốt 

Answer 3

\(A=x^2+y^2-x^2y-x^2y+2xy-5\)

\(=x^2+y^2-2x^2y+2xy-5\)

\(=\left(x+y\right)^2-xy\left(y+x\right)-5\)

\(=2^2-2xy-5=4-2xy-5=-1-2xy\)