Question

Biểu diễn các đa thức sau dưới dạng bình phương của 1 tổng:

a) x^2+2x(y+1)+y^2+2y+1

b) u^2+v^2+2u+2v+2(u+1)(v+1)+2

Answer 1

a, \(x^2+2x\left(y+1\right)+y^2+2y+1=\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1\)

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

b, \(u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+2\)

\(=u^2+v^2+2u+2v+2uv+2u+2v+2+2\)

\(=\left(u^2+2uv+v^2\right)+\left(4u+4v\right)+4\)

\(=\left(u+v\right)^2+4\left(u+v\right)+2^2=\left(u+v+2\right)^2\)

Answer 2

1.

a) \(A=x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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

b) \(B=u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+2\)\(B=u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+1+1\)\(B=\left(u^2+2u+1\right)+2\left(u+1\right)\left(v+1\right)+\left(v^2+2v+1\right)\)\(B=\left(u+1\right)^2+2\left(u+1\right)\left(v+1\right)+\left(v+1\right)^2\)\(B=\left(u+1+v+1\right)^2=\left(u+v+2\right)^2\)

tik mik nha !!!