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Ta có:
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)\)
\(=\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+c\right)\)(1)
\(b^2+ab-c^2-ac=\left(b^2-c^2\right)+\left(ab-ac\right)\)
\(=\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\)
\(=\left(b-c\right)\left(a+b+c\right)\)(2)
\(c^2+bc-a^2-ab=\left(c^2-a^2\right)+\left(bc-ab\right)\)
\(=\left(c-a\right)\left(a+c\right)+b\left(c-a\right)\)
\(=\left(c-a\right)\left(a+b+c\right)\)(3)
Ta có : \(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}\)\(+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}\)\(+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)(*)
Thế (1),(2),(3) vào (*)
=>\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Dễ thôi bạn chỉ cần quy đồng thôi
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\)\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
=\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
=\(\frac{c-a+a-b+b-c}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}=0\)
Ta có :\(\left(a-b\right)\left(c^2+bc-a^2-ab\right)=\left(a-b\right)\left[\left(c^2-a^2\right)+\left(bc-ab\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(a+b+c\right)\)
Tương tự : \(\left(b-c\right)\left(a^2+ac-b^2-bc\right)=\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)
\(\left(c-a\right)\left(b^2+ab-c^2-ac\right)=\left(c-a\right)\left(b-c\right)\left(a+b+c\right)\)
\(MTC=\left(a-b\right)\left(b-c\right)\left(c-s\right)\left(a+b+c\right)\)
Kí hiệu biểu thức đã cho bởi \(Q\),ta có :
\(Q=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Lời giải :
\(A=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{-bc\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-ac\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=\frac{-bc\left(b-c\right)-ac\left(c-a\right)-ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Xét tử số :
\(TS=-b^2c+bc^2-ac^2+a^2c-a^2b+ab^2\)
\(=-ab\left(a-b\right)-c^2\left(a-b\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(-ab-c^2+ac+bc\right)\)
\(=\left(a-b\right)\left[-a\left(b-c\right)+c\left(b-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Khi đó \(A=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
\(b.=\frac{1\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{1c-1a+1a-1b+1b-1c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{2b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(b-a\right)}\)
\(=-\frac{-c\left(a^2-b^2\right)+ab\left(a-b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-\frac{\left(a-b\right)\left[-c\left(a+b\right)+ab+c^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{\left(a-b\right)\left(-ac-bc+ab+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{-\left(a-b\right)\left[-b\left(c-a\right)+c\left(c-a\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{\left(a-b\right)\left(c-a\right)\left(-b+c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a-b\right)\left(c-a\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
Đặt
\(\Rightarrow\hept{\begin{cases}x=a-b\\y=a-c\\z=b-c\end{cases}}\)
Ta được
\(B=\frac{1}{axy}+\frac{1}{bxz}+\frac{1}{cyz}=\frac{bcz-acy+abx}{abcxyz}\)
\(=\frac{bc\left(b-c\right)-ac\left(a-c\right)+ab\left(a-b\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc\left(b-c\right)-ac\left(a-b+b-c\right)+ab\left(a-b\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc\left(b-c\right)-ac\left(a-b\right)-ac\left(b-c\right)+ab\left(a-b\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{c\left(b-c\right)\left(b-a\right)+a\left(a-b\right)\left(b-c\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)\left(a-b\right)\left(a-c\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{1)}{abc}\)
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