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1)
$2x=a+b+c$
$\Rightarrow x-a=\dfrac{b+c-a}{2},\quad x-b=\dfrac{c+a-b}{2},\quad x-c=\dfrac{a+b-c}{2}$
$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$
$=\dfrac{(b+c-a)(c+a-b)+(c+a-b)(a+b-c)+(a+b-c)(b+c-a)}{4}$
$=\dfrac{2ab+2bc-a^2-b^2-c^2+2bc+2ca-a^2-b^2-c^2+2ca+2ab-a^2-b^2-c^2}{4}$
$=\dfrac{4ab+4bc+4ca-3(a^2+b^2+c^2)}{4}$
$=\dfrac{(a+b+c)^2-2(a^2+b^2+c^2)}{4}$
$=\dfrac{(2x)^2-2(a^2+b^2+c^2)}{4}$
$=ab+bc+ca-x^2$
$\therefore\ (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=ab+ac+bc-x^2.$
2)
$ab+bc+ca=abc,\quad a+b+c=1$
$(a-1)(b-1)(c-1)$$=abc-ab-bc-ca+a+b+c-1$$=abc-ab-bc-ca+1-1$$=abc-(ab+bc+ca)$$=abc-abc$$=0$
$\therefore\ (a-1)(b-1)(c-1)=0.$
a: (x+a)(x+b)
\(=x^2+bx+ax+ab\)
\(=x^2+x\left(a+b\right)+ab\)
b: \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left(x^2+ax+bx+ab\right)\left(x+c\right)\)
\(=x^3+x^2c+ax^2+axc+bx^2+bxc+abx+abc\)
\(=x^3+x^2\left(a+b+c\right)+x\left(ab+bc+ca\right)+abc\)
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
$\dfrac{x-b-c}{a}+\dfrac{x-c-a}{b}+\dfrac{x-a-b}{c}=3$
$\Leftrightarrow x\left(\dfrac1a+\dfrac1b+\dfrac1c\right)-\left(\dfrac ba+\dfrac ca+\dfrac cb+\dfrac ab+\dfrac ac+\dfrac bc\right)=3$
$\Leftrightarrow x\left(\dfrac1a+\dfrac1b+\dfrac1c\right)=3+\dfrac ba+\dfrac ca+\dfrac cb+\dfrac ab+\dfrac ac+\dfrac bc$
$\Leftrightarrow x\cdot\dfrac{ab+bc+ca}{abc}=\dfrac{3abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2}{abc}$
$\Leftrightarrow x(ab+bc+ca)=3abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2$
$\Leftrightarrow x(ab+bc+ca)=3abc+(a+b+c)(ab+bc+ca)-3abc$
$\Leftrightarrow x(ab+bc+ca)=(a+b+c)(ab+bc+ca)$
$\Leftrightarrow x=a+b+c$
$\boxed{x=a+b+c}$
a. \(VT=\left(x+a\right)\left(x+b\right)=x^2+ã+bx+ab=x^2+\left(a+b\right)x+ab=VP\)
B. \(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left[\left(x+a\right)\left(x+b\right)\right].\left(x+c\right)\)
\(=\left[\left(x^2+\left(a+b\right)x\right)+ab\right].\left(x+c\right)=x^3+x^2c+\left(a+b\right)x^2+c\left(a+b\right)x+abx+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)
a ) VP = \(\left(x+a\right).\left(x+b\right)=x^2+bx+ax+ab\)
VT = \(x^2+\left(a+b\right).x+ab=x^2+ax+bx+ab\)
\(\Rightarrow VT=VP\)
b ) VP : \(\left(x+a\right).\left(x+b\right)\left(x+c\right)=\left(x^2+bx+ax+ab\right).\left(x+c\right)\) ( Vế đầu áp dụng luôn ở câu a )
\(=x^2.x+x^2.c+bx.x+bx.c+ax.x+ax.c+ab.x+ab.c\)
\(=x^3+cx^2+bx^2+cbx+ax^2+cax+abx+abc\)
\(=x^3+\left(cx^2+bx^2+ax^2\right)+\left(cbx+cax+abx\right)+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+ac+bc\right).x+abc\)
Vậy \(\left(x+a\right).\left(x+b\right).\left(x+c\right)=x^3+\left(a+b+c\right).x^2+\left(ab+ca+bc\right).x+abc\)
a) VP =\(\left(x+a\right)\left(x+b\right)=x^2+bx+\text{ax+ab}\)
\(VT=x^2+\left(a+b\right).x+ab=x^2+ax+bx+ab\\ =>VT=VP\)
b) VP : \(\left(x+a\right).\left(x+b\right).\left(x+c\right)=\left(x^2+bx+ax+ab\right).\left(x+c\right)\)( Vế đầu áp dụng luôn ở câu a )
\(=x^2.x+x^2.c+bx.x+bx.c+\text{ax}.x+\text{ax}.c+ab.c+ab.c\\ =x^3+cx^2+bx^2-cbx+\text{ax}^2+ca.x+ab.x+abc\\ \)
\(=x^3+\left(cx^2+bx^2+\text{ax}^2\right)-\left(cbx+c\text{ax}+abx\right)+abc\\ =x^3-\left(a+b+c\right)x^2+\left(ab+ac+bc\right).x+abc\)
Vậy \(\left(x+a\right)\left(x-b\right)\left(x+c\right)=x^3+\left(a+b+c\right).x^2+\left(ab+ca+bc\right).x+abc\)