(a+b)(b+c)(c+a) +abc= (a+b).(b.c + b.a + c.c + c.a) +abc
= (a+b).(a.b + b.c + c.a) + (a+b).(c.c) +abc
= (a+b+c).(a.b + b.c + c.a) - c.(a.b + b.c + c.a) + (a+b).(c.c) +abc
= (a+b+c).(a.b + b.c + c.a) - a.b.c - b.c.c - c.c.a + a.c.c + b.c.c +abc
= (a+b+c).(a.b + b.c + c.a) - a.b.c+abc
=(a+b+c)0+0=0
Bài này dễ thôi!
(a+b)(b+c)(c+a)+abc=(ab+ac+b2+bc)(c+a) +abc=b2(c+a) +abc= b2c + b2a + abc = b(bc+ab+ca) = b.0=0
ab+bc+ca=0 ab+bc+ca=0 ab+bc+ca=0
=>(ab+bc)+ca=0 =>(ab+ca)+bc=0 =>(bc+ca)+ab=0
=>b.(c+a)+ca=0 =>a.(b+c)+bc=0 =>c.(a+b)+ab=0
=>c+a=-ca/b =>b+c=-bc/a =>a+b=-ab/c
Suy ra: \(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc=\frac{-ab}{c}.\frac{-bc}{a}.\frac{-ca}{b}+abc\)
\(=\frac{-a^2b^2c^2}{abc}+abc=\frac{-a^2b^2c^2}{abc}+\frac{a^2b^2c^2}{abc}=\frac{0}{abc}=0\)
สาวแว่น อร้ายยย ......................................................................................................
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)-abc\)
\(=\left(ab+ac+b^2+bc\right)\left(c+a\right)+abc\)
\(=b^2\left(c+a\right)+abc=b^2c+ab^2+abc=b\left(bc+ab+ac\right)=0\)
