Question

Cho các số a,b,c,d thỏa mãn: abcd=1. Tính gt bthức:

\(\frac{1}{1+2a+3ab+4abc}+\frac{2}{2+3b+4bc+bcd}+\frac{3}{3+4c+cd+2acd}+\frac{4}{4+d+2ad+abd}\)

Giúp mình với...!!!

Answer 1

Nhầm, cái cuối là \(\frac{4}{4+d+2ad+3abd}\)

Answer 2

\(\frac{1}{1+2a+3ab+4abc}+\frac{2}{2+3b+4bc+bcd}+\frac{3}{3+4c+cd+2acd}+\frac{4}{4+d+2ad+3abd}\)

= \(\frac{1}{1+2a+3ab+4abc}+\frac{2a}{2a+3ab+4abc+abcd}+\frac{3ab}{3ab+4abc+abcd+2abacd}\)

\(+\frac{4abc}{4abc+abcd+2aabcd+3abcabd}\)

= \(\frac{1}{1+2a+3ab+4abc}+\frac{2a}{2a+3ab+4abc+1}+\frac{3ab}{3ab+4abc+1+2a}+\frac{4abc}{4abc+1+2a+3ab}\)

= \(\frac{1+2a+3ab+4abc}{1+2a+3ab+4abc}=1\)