Question

Cho a > b > 0 và a² - 6b² = -ab. Tính M = \(\frac{2ab}{2a^2-3b^2}\)

Answer 1

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

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

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

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

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

*)Xét \(a=-3b\) thay vào M ta có:

\(M=\frac{2\cdot3\left(-b\right)\cdot b}{2\left(-3b\right)^2-3b^2}=\frac{-6b^2}{15b^2}=-\frac{2}{5}\)

*)Xét \(a=2b\) thay vào M ta có:

\(M=\frac{2\cdot2b\cdot b}{2\cdot\left(2b\right)^2-3b^2}=\frac{4b^2}{8b^2-3b^2}=\frac{4b^2}{5b^2}=\frac{4}{5}\)