Ta có \(a.b=\frac{1}{2}\)\(=>\)\(a=\frac{1}{2b}\)(1)
Thay (1) vào \(ac=\frac{3}{4}\) ta có \(\frac{1}{2}b.c=\frac{3}{4}\)\(=>\)\(c=\frac{3b}{2}\)(2)
Thay (2) vào \(bc=\frac{2}{3}\) ta có \(b.\frac{3b}{2}=\frac{2}{3}\)\(=>\)\(b^2=\frac{2}{3}.\frac{2}{3}=\frac{4}{9}\)
\(=>\)b={\(\frac{-2}{3};\frac{2}{3}\)}
\(=>\)a={\(\frac{1}{2.\frac{-2}{3}}\);\(\frac{1}{2.\frac{2}{3}}\)}\(=> \)\(a=\left\{\frac{-3}{4};\frac{3}{4}\right\}\)
c=\(\left\{\frac{3.\frac{-2}{3}}{2};\frac{3.\frac{2}{3}}{2}\right\}=\left\{-1;1\right\}\)
Vậy a=\(\frac{-3}{4}\);b=\(\frac{-2}{3}\);c=\(-1\)
và a=\(\frac{3}{4};b=\frac{2}{3};c=1\)
ab=\(\frac{1}{2}\);bc=\(\frac{2}{3}\);ac=\(\frac{3}{4}\)
Nhân từng vế các đẳng thức trên ta được:
ab.bc.ac=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\)\(\Rightarrow\)(abc)2=\(\frac{1}{4}\)\(\Rightarrow\)abc=\(\frac{1}{2}\) hoặc abc=\(\frac{-1}{2}\)
+)abc=\(\frac{1}{2}\)
có ab=\(\frac{1}{2}\)\(\Rightarrow\)c=abc:ab=\(\frac{1}{2}:\frac{1}{2}=1\)
có bc=\(\frac{2}{3}\Rightarrow a=abc:bc=\frac{1}{2}:\frac{2}{3}=\frac{3}{4}\)
có ac=\(\frac{3}{4}\Rightarrow b=abc:ac=\frac{1}{2}:\frac{3}{4}=\frac{2}{3}\)
+)abc=-\(\frac{1}{2}\)xét tương tự abc=\(\frac{1}{2}:a=-\frac{3}{4};b=\frac{-2}{3};c=-1\)
vậy (a;b;c)={(\(\frac{3}{4};\frac{2}{3};1\));(\(\frac{-3}{4};\frac{-2}{3};-1\))}
