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15 tháng 6 2017

fgdcxynfyjfyj

15 tháng 6 2017

jffxhn

16 tháng 9 2018

Đặt 

\(\Rightarrow\hept{\begin{cases}x=a-b\\y=a-c\\z=b-c\end{cases}}\)

Ta được

\(B=\frac{1}{axy}+\frac{1}{bxz}+\frac{1}{cyz}=\frac{bcz-acy+abx}{abcxyz}\)

\(=\frac{bc\left(b-c\right)-ac\left(a-c\right)+ab\left(a-b\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{bc\left(b-c\right)-ac\left(a-b+b-c\right)+ab\left(a-b\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{bc\left(b-c\right)-ac\left(a-b\right)-ac\left(b-c\right)+ab\left(a-b\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{c\left(b-c\right)\left(b-a\right)+a\left(a-b\right)\left(b-c\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(b-c\right)\left(a-b\right)\left(a-c\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{1)}{abc}\)

Vậy ...

8 tháng 7 2016

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 (*)

=>\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)

\(\Leftrightarrow\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)

8 tháng 7 2016

Dễ thôi bạn chỉ cần quy đồng thôi

\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\)\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)

=\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)

=\(\frac{c-a+a-b+b-c}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}=0\)

8 tháng 7 2016

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 :

         \(Q=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)

20 tháng 2 2019

Ta có:\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(b-a\right)+\left(a-c\right)}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}+\frac{1}{c-a}\)

Chứng minh tương tự,ta được:

\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{a-b}+\frac{1}{b-c}\)

\(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{b-c}+\frac{1}{c-a}\)

\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)\left(đpcm\right)\)

1 tháng 2 2020

Ta có :

\(VT=\frac{1}{2}\left[\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}\right]\)

\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2}{\left(a-b\right)\left(a-c\right)}+\frac{\left(a-c\right)^2}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\frac{\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)

\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2+\left(a-c\right)^2+\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)

\(=\frac{1}{2}\left[\frac{b^2-2bc+c^2+a^2-2ac+c^2+a^2-2ab+b^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)

\(=\frac{1}{2}\left[\frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)

\(=\frac{a^2+b^2+c^2-ab-bc-ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)(1)

Lại có :

\(VP=\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)

\(=\frac{\left(b-c\right)\left(a-c\right)+\left(a-b\right)\left(a-c\right)-\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{ab-bc-ac+c^2+a^2-ac-ab+bc-ab+ac+b^2-bc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{a^2+b^2+c^2-ab-ac-bc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)(2)

Từ (1) và (2) \(\RightarrowĐPCM\)