Question

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

\(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

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

Viết biêu thức sau dưới dang một tổng hoac một hiệu

Answer 1

x²  - 2x(y+2) + y² + 4x + 4

= x² - 2x(y+2) + ( y+2)²

= (x - y + 2)²

Answer 2

\(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x+3\right)\left(x+6\right)\left(x+4\right)\left(x+5\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)(1)

Đặt \(x^2+9x+18=a\)

\(\Rightarrow\left(1\right)=a\left(a+2\right)+1=a^2+2a+1=\left(a+1\right)^2\)

\(=\left(x^2+9x+19\right)^2\)

Answer 3

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

\(=y^2+x^2+4-2xy-4x+4y\)

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

Answer 4

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

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

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

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