Question

cho a,b > 0 thỏa mãn: 

\(a^2+ab-6b^2=0\)

Tính: \(S=\frac{a+3b}{5a+b}\)

Answer 1

Từ \(a^2+ab-6b^2=0\Rightarrow\left(a^2+3ab\right)-\left(2ab+6b^2\right)=0\)

\(\Leftrightarrow a\left(a+3b\right)-2b\left(a+3b\right)=0\Leftrightarrow\left(a+3b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-3b\\a=2b\end{cases}}\)

Với \(a=-3b\Rightarrow S=\frac{-3b+3b}{5.\left(-3b\right)+b}=\frac{0}{-14b}=0\)

Với \(a=2b\Rightarrow S=\frac{2b+3b}{5.2b+b}=\frac{5b}{11b}=\frac{5}{11}\)