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28 tháng 10 2021

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

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

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

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

28 tháng 10 2021

\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

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

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

28 tháng 10 2021

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

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

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

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

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

 

NV
20 tháng 12 2020

\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)

\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

\(=2\left(a+b+c\right)\)

2 tháng 3 2017

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

<=>A=\(\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

<=> A=\(\dfrac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)<=> A=0

2 tháng 3 2017

Đặt: \(a-b=x\)

\(a-c=y\)

\(b-c=z\)

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

\(=\dfrac{1}{xy}-\dfrac{1}{xz}+\dfrac{1}{yz}\)

\(=\dfrac{xyz^2-xy^2z+x^2yz}{x^2y^2z^2}\)

\(=\dfrac{xyz\left(z-y+x\right)}{x^2y^2z^2}\)

\(=\dfrac{z-y+x}{xyz}\)

Thay \(a-b=x;a-c=y;b-c=z\) vào biểu thức \(\dfrac{z-y+x}{xyz}\), ta được:

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

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

= 0

Vậy:\(A=\dfrac{1}{\left(a-b\right)\left(a-c\right)}+\dfrac{1}{\left(b-a\right)\left(b-c\right)}+\dfrac{1}{\left(c-a\right)\left(c-b\right)}=0\)

17 tháng 12 2020

Ta có: \(A=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)

\(=\dfrac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\dfrac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\dfrac{c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

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

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

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

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

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

10 tháng 7 2017

\(T=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)

Do a+b+c =0 nên => a+b = (-c) => \(\left(a+b\right)^2=\left(-c\right)^2=>a^2+2ab+b^2=c^2\)

\(=>a^2+b^2-c^2=-2ab\)

Làm tương tự trên ta có : \(b^2-c^2-a^2=2ac;\)

\(a^2-b^2-c^2=2bc;\)

\(=>T=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}\)

Với a+b+c = 0 thì \(a^3+b^3+c^3=3abc\) (bạn tự chứng minh hằng đẳng thức mở rộng nhé);

\(=>T=\dfrac{3abc}{2abc}=\dfrac{3}{2}=1,5\)

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