Ta có: VT=(x-a).(x-b)+(x-b).(x-c)+(x-c).(x-a)
=x2-ax-bx+ab+x2-bx-cx+bc+x2-cx-ax+ca
=3.x2-2.(ax+bx+cx)+ab+bc+ca
=3.x2-2x.(a+b+c)+ab+bc+ca
=x.[3x-2.(a+b+c)]+ab+bc+ca
Vì \(x=\frac{a+b+c}{2}\)
<=>a+b+c=2x
<=>2.(a+b+c)=4x
<=>3x-2.(a+b+c)=-x
=>VT=x.(-x)+ab+bc+ca
=ab+bc+ca-x2=VP
=>ĐPCM
Xét \(VT=\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)\)
\(=\left(x^2-ax-bx+ab\right)+\left(x^2-bx-cx+bc\right)+\left(x^2-ax-cx+ac\right)\)
\(=3x^2-2x\left(a+b+c\right)+ab+bc+ca \left(1\right)\)
từ \(x=\frac{1}{2}\left(a+b+c\right)=>2x=a+b+c\)
Do đó \(\left(1\right)=3x^2-2x.2x+ab+bc+ca=3x^2-4x^2+ab+bc+ca=ab+bc+ca-x^2=VP\left(đpcm\right)\)
TA có :
( x-a ) . ( x-b) + ( x - b ). ( x- c ) + (x-c).( x-a )
= x2 - bx - ax + ab + x2 - cx - bx + bc + x2 - ax - cx + ac
= 3x2 - ( 2ax + 2bx + 2cx )
= 3x2 - 2x ( a+b+c )
= 3x2 - \(2\frac{a+b+c}{2}\). ( a + b + c )
= 3x2
Lại có ab + bc + ca - x2
= ab + bc + ca - \(\left(\frac{a+b+c}{2}\right)^2=ab+bc+ca-\frac{a^2+b^2+c^2+2ab+2bc+2ca}{4}\)
= \(\frac{4ab+4bc+4ca}{4}-\frac{a^2+b^2+c^2+2ab+2ca+2bc}{4}\)
\(=\frac{2ab+2bc+2ca-a^2-b^2-c^2}{4}=3x^2\)
=> đpcm
cũng thank bạn nha Trịnh Tiến Đức nhưng chỉ cần xét 1 vế rồi suy ra cho nhanh chư cách bạn hơi dai nhưng m` vẫn hiểu cam ơn nha ~~~
