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\(a)\) ĐKXĐ: \(a\ne-b;a\ne-c;b\ne-c\)
\(\dfrac{x-ab}{a+b}+\dfrac{x-ac}{a+c}+\dfrac{x-bc}{b+c}=a+b+c\)
\(\Leftrightarrow\left(\dfrac{x-ab}{a+b}-c\right)+\left(\dfrac{x-ac}{a+c}-b\right)+\left(\dfrac{x-bc}{b+c}-a\right)=0\)
\(\Leftrightarrow\dfrac{x-ab-ac-bc}{a+b}+\dfrac{x-ac-ab-bc}{a+c}+\dfrac{x-bc-ab-ac}{b+c}=0\)
\(\Leftrightarrow\left(x-ab-ac-bc\right)\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)=0\)
Vì \(a,b,c>0\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}>0\)
\(\Leftrightarrow x-ab-ac-bc=0\)
\(\Leftrightarrow x=ab+ac+bc\)
a: (x+a)(x+b)
\(=x^2+bx+ax+ab\)
\(=x^2+x\left(a+b\right)+ab\)
b: \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left(x^2+ax+bx+ab\right)\left(x+c\right)\)
\(=x^3+x^2c+ax^2+axc+bx^2+bxc+abx+abc\)
\(=x^3+x^2\left(a+b+c\right)+x\left(ab+bc+ca\right)+abc\)
a,ta có : x^3+y^3-xy(x+y)=(x+y)(x^2+y^2-xy) -xy(x+y)=(x+y)(x^2+y^2-2xy=(x+y)(x-y)^2
(đpcm)
b)x^3-y^3+xy(x-y)=(x-y)(x^2+y^2+xy)+xy(x-y)=(x-y)(x^2+y^2+2xy)=(x-y)(x+y)^2 (đpcm)
P/s : Phần b ) : \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
a ) \(\left(x+a\right)\left(x+b\right)=x^2+ax+bx+ab=x^2+\left(a+b\right)x+ab\)
b ) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)
\(=x^2\left(x+c\right)+\left(a+b\right)x\left(x+c\right)+ab\left(x+c\right)\)
\(=x^3+x^2c+\left(ax+bx\right)\left(x+c\right)+abx+abc\)
\(=x^3+x^2c+ax^2+bx^2+axc+bxc+abx+abc\)
\(=x^3+\left(x^2a+x^2b+x^2c\right)+\left(abx+bcx+axc\right)+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
chịu!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
a3x-ab+b-x
=x(a3-1)-b(a-1)
=(a-1)(xa2+xa+x-b)