Ta có : \(ab.bc.ac=\) \(\dfrac{1}{2}\).\(\dfrac{2}{3}\).\(\dfrac{3}{4}\)
\(\left(abc\right)\left(abc\right)\)\(\dfrac{1}{4}\)
\(\left(abc\right)^2=\dfrac{1}{4}\)
\(\left(abc\right)^2=\left(\dfrac{1}{2}\right)^2\)
=> \(abc=\dfrac{1}{2}\)
Ta có : \(c=abc:ab=\dfrac{1}{2}:\dfrac{1}{2}=1\)
\(a=abc:bc=\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{3}{4}\)
\(b=abc:a:c=\dfrac{1}{2}:1:\dfrac{3}{4}=\dfrac{2}{3}\)
Vậy ....
Theo đề bài cho :
\(ab=\dfrac{1}{2}\) ; \(bc=\dfrac{2}{3}\) ; \(ac=\dfrac{3}{4}\)
Suy ra :
\(ab.bc.ac=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\)
\(\Rightarrow\left(abc\right)^2=\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^2\)
\(\Rightarrow abc=\dfrac{1}{2}\)
Mà \(ab=\dfrac{1}{2}\) nên \(c=1\)
Thay \(c=1\) ta được \(\left\{{}\begin{matrix}bc=\dfrac{2}{3}\\ac=\dfrac{3}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b=\dfrac{2}{3}\\a=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(a=\dfrac{3}{4};b=\dfrac{2}{3};c=1\)
cả 2 bạn hình như thiếu trường hợp thì phải
