Question

Cho \(3\)\(a^2\)+\(b^2=4ab\). Tính \(P=\dfrac{a-b}{a+b}\)

Answer 1

theo bài ra ta có:

\(3a^2+b^2=4ab\\ \Rightarrow3a^2+b^2-4ab=0\\ \Rightarrow4a^2-4ab+b^2-a^2=0\\ \Rightarrow\left(2a-b\right)^2-a^2=0\\ \Rightarrow\left(2a-b-a\right)\left(2a-b+a\right)=0\\ \Rightarrow\left(a-b\right)\left(3a-b\right)=0\\ \Rightarrow\left[{}\begin{matrix}a-b=0\\3a-b=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}a=b\\3a=b\end{matrix}\right.\)

TH1: Nếu a = b thì:

\(P=\dfrac{a-a}{a+a}=\dfrac{0}{2a}=0\)

vậy P = 0

TH2: Nếu 3a = b thì:

\(P=\dfrac{a-3a}{a+3a}=\dfrac{-2a}{4a}=\dfrac{-1}{2}\)

vậy \(P=\dfrac{-1}{2}\)