Question

Cho biểu thức :

P=\(\left(a+1\right)^2+\left(b+1\right)^2+\left(c+1\right)^2+2\left(ab+ac+bc\right)\)

Q=\(\left(a+b+c+1\right)^2\)

Tính P-Q

Answer 1

ta có : \(P=\left(a+1\right)^2+\left(b+1\right)^2+\left(c+1\right)^2+2\left(ab+bc+ca\right)\)

\(P=a^2+2a+1+b^2+2b+1+c^2+2c+1+2ab+2bc+2ca\)

\(P=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+2a+2b+2c+3\)

\(P=\left(a+b+c\right)^2+2\left(a+b+c\right)+3\)

ta có : \(Q=\left(a+b+c+1\right)^2=\left(\left(a+b+c\right)+1\right)^2\)

\(Q=\left(a+b+c\right)^2+2\left(a+b+c\right)+1\)

\(\Leftrightarrow P-Q=\left(a+b+c\right)^2+2\left(a+b+c\right)+3-\left(a+b+c\right)^2-2\left(a+b+c\right)-1=2\)

vậy \(P-Q=2\)