Violympic toán 8

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CHO abcd=1. Tính:

A=\(\dfrac{a}{abc+ab+a+1}+\dfrac{b}{bcd+bc+b+1}+\dfrac{c}{cda+cd+c+1}+\dfrac{d}{dab+da+d+1}\)

Nguyễn Việt Lâm
27 tháng 11 2018 lúc 9:36

\(A=\dfrac{a}{abc+ab+a+1}+\dfrac{ba}{abcd+abc+ab+a}+\dfrac{\dfrac{c}{cd}}{\dfrac{acd}{cd}+\dfrac{cd}{cd}+\dfrac{c}{cd}+\dfrac{1}{cd}}+\dfrac{\dfrac{d}{d}}{\dfrac{dab}{d}+\dfrac{ad}{d}+\dfrac{d}{d}+\dfrac{1}{d}}\)

\(A=\dfrac{a}{abc+ab+a+1}+\dfrac{ab}{1+abc+ab+a}+\dfrac{\dfrac{1}{d}}{a+1+\dfrac{1}{d}+\dfrac{1}{cd}}+\dfrac{1}{ab+a+1+\dfrac{1}{d}}\)

\(abcd=1\Rightarrow\dfrac{1}{d}=abc;\dfrac{1}{cd}=ab\)

\(\Rightarrow A=\dfrac{a}{abc+ab+a+a}+\dfrac{ab}{abc+ab+a+1}+\dfrac{abc}{a+1+abc+ab}+\dfrac{1}{ab+a+1+abc}\)

\(\Rightarrow A=\dfrac{a+ab+abc+1}{abc+ab+a+1}=1\)