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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
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 \(\left(x+a\right)\left(x-b\right)\left(x+c\right)=x^3+\left(a+b+c\right).x^2+\left(ab+ca+bc\right).x+abc\)