Giải thích các bước giải:
Ta có:
$B=\dfrac{1}{abc+ab+a+1}+\dfrac{1}{bcd+bc+b+1}+\dfrac{1}{cda+cd+c+1}+\dfrac{1}{abd+ad+d+1}$
$\to B=\dfrac{1}{abc+ab+a+1}+\dfrac{a}{a(bcd+bc+b+1)}+\dfrac{ab}{ab(cda+cd+c+1)}+\dfrac{abc}{abc(abd+ad+d+1)}$
$\to B=\dfrac{1}{abc+ab+a+1}+\dfrac{a}{abcd+abc+ab+a}+\dfrac{ab}{abcda+abcd+abc+ab}+\dfrac{abc}{a^2b^2cd+a^2bcd+abcd+abc}$
$\to B=\dfrac{1}{abc+ab+a+1}+\dfrac{a}{1+abc+ab+a}+\dfrac{ab}{a+1+abc+ab}+\dfrac{abc}{ab+a+1+abc}$ vì $abcd=1$
$\to B=\dfrac{1+a+ab+abc}{abc+ab+a+1}$
$\to B=1$
