Question

Biết \(a^3-3ab^2=5\) và  \(b^3-3a^2b=10\)

TÍNH\(\frac{a^2+b^2}{2018}\)

Answer 1

Ta có : \(a^3-3ab^2=5\Rightarrow\left(a^3-3ab^2\right)^2\)\(=25\Rightarrow a^6-6a^4b^2+9a^2b^4=25\)

            \(b^3-3a^2b=10\Rightarrow\left(b^3-3a^2b\right)^2=100\)\(\Rightarrow b^6-6a^2b^4+9a^4b^2=100\)

Cộng hai vế ta được : 

\(a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=125\)

\(\Rightarrow a^6+3a^4b^2+3a^2b^4+b^6=125\)

\(\Rightarrow\left(a^2+b^2\right)^3=125\)

\(\Rightarrow\left(a^2+b^2\right)^3=5^3\)

\(\Rightarrow a^2+b^2=5\)

\(\Rightarrow\frac{a^2+b^2}{2018}=\frac{5}{2018}\)

Chúc bạn học tốt ^^

Answer 2

what đè he