Ta có: \(\left\{{}\begin{matrix}a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(c+a\right)\\b^2+1=b^2+ab+bc+ca=b\left(b+c\right)+a\left(b+c\right)=\left(b+c\right)\left(a+b\right)\\c^2+1=c^2+ab+bc+ca=c\left(c+a\right)+b\left(c+a\right)=\left(c+a\right)\left(b+c\right)\end{matrix}\right.\)
- Từ đây suy ra:
\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
- Vì \(a,b,c\in Q\), nên \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\) là bình phương của 1 số hữu tỉ (đpcm).
Ta có:
( a2 + 1) ( b2 + 1) ( c2 + 1)
= ( a2 + ab + bc + ca) ( b2 + ab + bc + ca) ( c2 + ab + bc + ca)
= [(a + ab) + ( bc + ca)] [(b2 + ab) + ( bc + ca)] [( c2 + bc) + ( ab + ca)]
= (a + c ) ( a + b) (a + b) (b + c) (c + a) (b + c)
= [ ( a + b ) ( b + c) ( c + a)]2
Tích cho mình nha
