Question

Cho hai số dương a, b thỏa mãn \(log_4\left(2a+3b\right)=log_{10}a=log_{25}b\). Tính giá trị của biểu thức \(P=\dfrac{a^3-ab^2+b^3}{a^3+ab^2-3b^3}\)

Answer 1

Đặt \(log_4\left(2a+3b\right)=log_{10}a=log_{25}b=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=10^k\\b=25^k\\2a+3b=4^k\end{matrix}\right.\) \(\Rightarrow2.10^k+3.25^k=4^k\)

\(\Rightarrow2.\left(\dfrac{5}{2}\right)^k+3.\left(\dfrac{25}{4}\right)^k=1\)

Đặt \(\left(\dfrac{5}{2}\right)^k=x>0\)

\(\Rightarrow3x^2+2x-1=0\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left(\dfrac{5}{2}\right)^k=\dfrac{1}{3}\) \(\Rightarrow\dfrac{b}{a}=\left(\dfrac{25}{10}\right)^k=\left(\dfrac{5}{2}\right)^k=\dfrac{1}{3}\)

\(P=\dfrac{1-\left(\dfrac{b}{a}\right)^2+\left(\dfrac{b}{a}\right)^3}{1+\left(\dfrac{b}{a}\right)^2-3\left(\dfrac{b}{a}\right)^3}=...\)