Question

Cho a >b > 0 và 2(a² +b²) = 5ab

Tính P=\(\frac{3a-b}{2a+b}\)

Answer 1

\(2\left(a^2+b^2\right)=5ab\Leftrightarrow2a^2-5ab+2b^2=0\)

\(\Leftrightarrow2a^2-4ab-ab+2b^2=0\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\)

TH1: \(2a=b\Rightarrow P=\frac{3a-2a}{2a+2a}=\frac{a}{4a}=\frac{1}{4}\)

TH2: \(a=2b\Rightarrow P=\frac{6b-b}{4b+b}=\frac{5b}{5b}=1\)