Question

Cho a>b>c thỏa mãn : 3a^6+3.b^2 = 10ab

Tính P=\(\frac{a-b}{a+b}\)

Answer 1

\(3a^2+3b^2=10ab\)

\(\Rightarrow3a^2-10ab+3b^2=0\)

\(\Rightarrow3a^2-9ab-ab+3b^2=0\)

\(\Rightarrow3a\left(a-3b\right)-b\left(a-3b\right)=0\)

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

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

\(a>b>0\)

\(\Rightarrow a=3b\)

Thay vào biểu thức ta có:

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