Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(a^3-3ab^2\right)^2=25\Leftrightarrow a^6-6a^4b^2+9a^2b^4=25\)
\(\left(b^3-3a^2b\right)^2=100\Leftrightarrow b^6-6a^2b^4+9a^4b^2=100\)
\(\Rightarrow a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=125\)
\(\Leftrightarrow\left(a^2+b^2\right)^2=125\Leftrightarrow a^2+b^2=5\)
Thay a2+b2=5 vào S=2018a2+2018b2=2018(a2+b2)=2018.5=10090
ta có: (a3-3ab2)2=a6-6a4b2+9a2b4=25
(b3-3a2b)2=b6-6a2b4+9a4b2=100
=> (a3-3ab2)2-(b3-3a2b)2=a6-6a4b2+9a2b4+b6-6a2b4+9a4b2=125
<=>a6+3a4b2+3a2b4+b6=125
<=>(a2+b2)3=125
=>a2+b2=5
\(a^3-3ab^2=5=>(a^3-3ab^2)^2=25\)
\(b^3-3a^2b=10=>(b^3-3a^2b)^2=100\)
=>\(a^6-6a^4b^2+9a^2b^4\)=25
\(b^6-6a^2b^4+9a^4b^2=100\)
=>\(a^6+3a^2b^4+3a^4b^2+b^6=125\)
=>(\(a^2+b^2)^3=125\)
=>\(a^2+b^2=5\)
=>2016\(a^2+2016b^2=10080\)
Ta có :
+) \(a^3-3ab^2=5\Leftrightarrow\left(a^3-3ab^2\right)^2=25\Leftrightarrow a^6-6a^4b^2+9a^2b^4=25\)
+) \(b^3-3a^2b=10\Leftrightarrow\left(b^3-3a^2b\right)^2=100\Leftrightarrow b^6-6a^2b^4+9a^4b^2=100\)
\(\Leftrightarrow a^6+b^6+3a^2b^4+3a^4b^2=125\)
\(\Leftrightarrow\left(a^2+b^2\right)^3=125\)
\(\Leftrightarrow a^2+b^2=5\)
Ta cos :
\(S=2018a^2+2018b^2=2018\left(a^2+b^2\right)=2018.5=10090\)
Vaayj...