cho abcd=1 tinh S bt S = a/abc+ab+a+1+b/bcd+bc+b+1+c/cda+cd+c+1+d/dab+da+d+1 14/07/2021 Bởi Everleigh cho abcd=1 tinh S bt S = a/abc+ab+a+1+b/bcd+bc+b+1+c/cda+cd+c+1+d/dab+da+d+1
Bài giải: Ta có: S$=\frac{a}{abc}+ab+a+1+\frac{b}{bcd}+bc+b+1+\frac{c}{cda}+cd+c+1+\frac{d}{dab}+da+d+1$ $=4+a+b+c+d+ab+bc+cd+da+\frac{a}{abc}+\frac{b}{bcd}+\frac{c}{cda}+\frac{d}{dab}$ $=\frac{4abcd+a^2bcd+ab^2cd+abc^2d+abcd^2+a^2b^2cd+ab^2c^2d+abc^2d^2+a^2bcd^2+ad+ba+cb+dc}{abcd}$ $=\frac{abcd.(4+a+b+c+d+ab+bc+cd+da)+ad+ba+cb+dc}{abcd}$ $=4+a+b+c+d+ab+bc+cd+da+ad+ba+cb+dc$ $=4+a+b+c+d+2.(ab+bc+cd+da)$ Bình luận
Bài giải:
Ta có:
S$=\frac{a}{abc}+ab+a+1+\frac{b}{bcd}+bc+b+1+\frac{c}{cda}+cd+c+1+\frac{d}{dab}+da+d+1$
$=4+a+b+c+d+ab+bc+cd+da+\frac{a}{abc}+\frac{b}{bcd}+\frac{c}{cda}+\frac{d}{dab}$
$=\frac{4abcd+a^2bcd+ab^2cd+abc^2d+abcd^2+a^2b^2cd+ab^2c^2d+abc^2d^2+a^2bcd^2+ad+ba+cb+dc}{abcd}$
$=\frac{abcd.(4+a+b+c+d+ab+bc+cd+da)+ad+ba+cb+dc}{abcd}$
$=4+a+b+c+d+ab+bc+cd+da+ad+ba+cb+dc$
$=4+a+b+c+d+2.(ab+bc+cd+da)$