\(a\cdot b=a:b\)
\(b\cdot b=a:a\)
\(b^2=1\)
\(b=\pm\sqrt{1}\)
\(b=\pm1\)
\(\orbr{\begin{cases}b=1\\b=-1\end{cases}}\)
TH 1 :
\(b=1\)
\(4\left(a-b\right)=a\cdot b\)
\(4\left(a-1\right)=a\cdot1\)
\(4a-4=a\)
\(3a=4\)
\(a=\frac{4}{3}\)
TH 2 :
\(b=-1\)
\(4\left(a-b\right)=a\cdot b\)
\(4\left(a-\left(-1\right)\right)=a\cdot\left(-1\right)\)
\(4\left(a+1\right)=-a\)
\(4a+4=-a\)
\(5a=-4\)
\(a=\frac{-4}{5}\)
Vậy \(\hept{\begin{cases}a=\frac{4}{3}\\b=1\end{cases}}\) hoặc \(\hept{\begin{cases}a=\frac{-4}{5}\\b=-1\end{cases}}\)
Bài giải
\(4\left(a-b\right)=ab=\frac{a}{b}\)
\(4a-4b=ab=\frac{a}{b}\)
Vì \(ab=\frac{a}{b}\text{ }\Rightarrow\text{ }ab^2=a\text{ }\Rightarrow\text{ }b^2=1\text{ }\Rightarrow\text{ }b=\pm1\)
TH 1 ; Với a = - 1 thì :
\(\Rightarrow\text{ }-4-4b=-b\text{ }\Rightarrow\text{ }-4=3b\text{ }\Rightarrow\text{ }b=-\frac{4}{3}\)
TH 2 : Với a = 1 thì :
\(\Rightarrow\text{ }4-4b=b\text{ }\Rightarrow\text{ }4=5b\text{ }\Rightarrow\text{ }b=\frac{4}{5}\)
Vậy ...
+ Nếu: a = 0
=> \(4\left(0-b\right)=0\)
\(\Leftrightarrow-4b=0\Rightarrow b=0\)
=> a = b = 0 không thỏa mãn vì b khác 0
+ Nếu a khác 0
Ta có: \(a\cdot b=\frac{a}{b}\Leftrightarrow b^2=1\Rightarrow\orbr{\begin{cases}b=1\\b=-1\end{cases}}\)
Xét b = 1: \(4\left(a-1\right)=a\)
\(\Leftrightarrow4a-4=a\)
\(\Leftrightarrow3a=4\)
\(\Rightarrow a=\frac{4}{3}\)
Xét b = -1: \(4\left(a+1\right)=-a\)
\(\Leftrightarrow4a+4=-a\)
\(\Leftrightarrow5a=-4\)
\(\Rightarrow a=-\frac{4}{5}\)
Vậy ta có 2 cặp số (a;b) thỏa mãn: \(\left(\frac{4}{3};1\right);\left(-\frac{4}{5};-1\right)\)
