Tìm a, b biết a-b=2(a+b)=a/b

a) Theo đề bài, ta có :

\(a-b=2\left(a+b\right)=\frac{a}{b}\\ \Leftrightarrow a-b=2a+2b\\ \Leftrightarrow a-2a=b+2b\\ \Leftrightarrow-a=3b\\ \Leftrightarrow a=-3b\)

Thay a = -3b vào \(a-b=\frac{a}{b}\), ta được :

\(-3b-b=-\frac{3b}{b}\\ \Leftrightarrow-4b=-3\\ \Leftrightarrow b=-\frac{3}{-4}=\frac{3}{4}\)

Vì :

\(a=-3b\\ \Rightarrow a=-3\cdot\frac{3}{4}=-\frac{9}{4}\)

Vậy :

\(\left\{\begin{matrix}a=-\frac{9}{4}\\b=\frac{3}{4}\end{matrix}\right.\)

b) Theo đề bài, ta có :

\(a+b=ab=\frac{a}{b}\\ \Rightarrow a=ab^2\\ \Rightarrow b^2=\frac{a}{a}=1\\ \Rightarrow\left[\begin{matrix}b=1\\b=-1\end{matrix}\right.\)

TH₁ : b = 1

\(\Rightarrow a+1=a\cdot1\\ \Rightarrow a+1=a\\ \Rightarrow a-a=1\)

\(\Rightarrow0=1\) ( Vô lý )

TH₂ : \(b=-1\)

\(\Rightarrow a-1=a\cdot\left(-1\right)\\ \Rightarrow a-1=-a\\ \Rightarrow2a=1\\ \Rightarrow a=\frac{1}{2}\)

Vậy :

\(\left\{\begin{matrix}a=\frac{1}{2}\\b=-1\end{matrix}\right.\)

c) Theo đề bài, ta có :

\(\left\{\begin{matrix}ab=2\\bc=3\\ac=54\end{matrix}\right.\)

\(\Rightarrow\frac{b}{c}=\frac{ab}{ac}=\frac{2}{54}=\frac{1}{27}\\ \Rightarrow\frac{b}{1}=\frac{c}{27}\\ \Rightarrow\frac{b^2}{1}=\frac{c^2}{729}=\frac{bc}{27\cdot1}=\frac{3}{27}=\frac{1}{9}\)

\(\Rightarrow\left\{\begin{matrix}b^2=\frac{1}{9}\cdot1=\frac{1}{9}\\c^2=\frac{1}{9}\cdot729=81\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}b=\sqrt{\frac{1}{9}}=\frac{1}{3}\\c=\sqrt{81}=9\end{matrix}\right.\)

Vì \(\left\{\begin{matrix}ac=54\\c=91\end{matrix}\right.\)

\(\Rightarrow a=\frac{54}{9}=6\)

Vậy :

\(\left\{\begin{matrix}a=6\\b=\frac{1}{3}\\c=9\end{matrix}\right.\)