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a) sửa đề: \(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
=\(\frac{-x^2\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-y^2\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(-\frac{\left\lbrace x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
xét tử số:
Tử=\(x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)
=\(x^2\left(y-z\right)-x\left(y^2-z^2\right)+yz\left(y-z\right)\)
=\(x^2\left(y-z\right)-x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\)
=\(\left(y-z\right)\left\lbrace x^2-x\left(y+z\right)+yz\right\rbrace\)
=\(\left(y-z\right)\left\lbrace x\left(x-y\right)-z\left(x-y\right)\right\rbrace\)
=\(\left(y-z\right)\left(x-y\right)\left(x-z\right)\)
=\(-\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
thay lại vào biểu thức cũ:
\(\Rightarrow-\frac{\left\lbrace-\left(x-y\right)\left(y-z\right)\left(z-x\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(1\)
b) \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)
=\(\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(P=x^2-x\left(a+b\right)+ab+x^2-x\left(b+c\right)+bc+x^2-x\left(c+a\right)+ac+x^2\)
\(=4x^2-2x\left(a+b+c\right)+\left(ab+bc+ac\right)\)
Thay x được \(P=\left(a+b+c\right)^2-\left(a+b+c\right)^2+\left(ab+bc+ca\right)=ab+bc+ac\)