Cho a,b,c tm: b-c/(a-b)(a-c) + c-a/(b-a)(b-c) + a-b/(c-a)(c-b) = 2015 Tính gtrị biểu thức: A= 1/a-b + 1/b-c + 1/c-a

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1. Ta có:

   `\qquad {b-c}/{(a-b)(a-c)} `

   `={(a-c)-(a-b)}/{(a-b)(a-c)}`

   `={a-c}/{(a-b)(a-c)}-{a-b}/{(a-b)(a-c)}`

   `=1/{a-b}-1/{a-c}=1/{a-b}+1/{c-a}`

   $\\$

   `\qquad {c-a}/{(b-a)(b-c)} `

   `={(b-a)-(b-c)}/{(b-a)(b-c)}`

   `={b-a}/{(b-a)(b-c)}-{b-c}/{(b-a)(b-c)}`

   `=1/{b-c}-1/{b-a}=1/{b-c}+1/{a-b}`

   $\\$

   `\qquad {a-b}/{(c-a)(c-b)}`

   `={(c-b)-(c-a)}/{(c-a)(c-b)}`

   `={c-b}/{(c-a)(c-b)}-{c-a}/{(c-a)(c-b)}`

   `=1/{c-a}-1/{c-b}=1/{c-a}+1/{b-c}`

   $\\$

   Theo đề bài:

   `{b-c}/{(a-b)(a-c)}  +  {c-a}/{(b-a)(b-c)} +   {a-b}/{(c-a)(c-b)}  = 2015`

   `=>1/{a-b}+1/{c-a}+1/{b-c}+1/{a-b}+1/{c-a}+1/{b-c}=2015`

   `=>2/{a-b}+2/{b-c}+2/{c-a}=2015`

   `=>2(1/{a-b}+1/{b-c}+1/{c-a})=2015`

   `=>1/{a-b}+1/{b-c}+1/{c-a}={2015}/2`

   $\\$

   Vậy `A=1/{a-b}+1/{b-c}+1/{c-a}={2015}/2`

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