Question

Cho \(\frac{a-b}{a^2+ab}+\frac{a+b}{a^2-ab}=\frac{3a-b}{a^2-b^2}\)

Tính P = \(\frac{a^3+2a^2b+2b^3}{2a^3+ab^2+2b^3}\)

Answer 1

Coi như biểu thức xác định

\(\frac{a-b}{a\left(a+b\right)}+\frac{a+b}{a\left(a-b\right)}=\frac{3a-b}{\left(a-b\right)\left(a+b\right)}\)

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

\(\Leftrightarrow2a^2+2b^2=3a^2-ab\)

\(\Leftrightarrow a^2-ab-2b^2=0\)

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

\(\Leftrightarrow a=2b\Leftrightarrow\frac{a}{b}=2\)

\(P=\frac{\left(\frac{a}{b}\right)^3+2\left(\frac{a}{b}\right)^2+2}{2\left(\frac{a}{b}\right)^3+\frac{a}{b}+2}=\frac{2^3+2.2^2+2}{2.2^3+2+2}=...\)