Question

Answer 1

1.\(A=x^2+y^2+2xy+2x+3y-5\)

\(=\left(x+y\right)^2+2x+3y-5\)

Thay x=1,y=2 vào biểu thức A ta có:

\(\left(x+y\right)^2+2x+3y-5\)

\(=\left(1+2\right)^3+2.1+3.2-5\)

\(=27+2+6-5\)

\(=30\)

Answer 2

lm hết hộ mk vs

mk cảm ơn rất nhiều

 

Answer 3

2.\(M=x^3+y^3+3xy\)

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

\(=x^2-xy+y^2+3xy\) (vì x+y=1)

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

\(=\left(x+y\right)^2=1\) (vì x+y=1)

Answer 4

3.\(A=x^2+2xy+y^2-4x-4y+1\)

\(=\left(x^2+2y+y^2\right)-\left(4x+4y\right)+1\)

\(=\left(x+y\right)^2-4\left(x+y\right)+1\)

Vì x+y=3, ta có:

\(\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=3^2-4.3+1\)

\(=9-12+1\)

\(=-2\)