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1) \(\left(a+b\right)^3=\left(a+b\right)\left(a+b\right)^2=\left(a+b\right)\left(a^2+2ab+b^2\right)\)
\(=a^3+2a^2b+ab^2+a^2b+2ab^2+b^3\)
\(=a^3+3a^2b+3ab^2+b^3\)
2) \(\left(a-b\right)^3=\left(a-b\right)\left(a-b\right)^2=\left(a-b\right)\left(a^2-2ab+b^2\right)\)\(=a^3-2a^2b+ab^2-a^2b+2ab^2-b^3\)
\(=a^3-3a^2b+3ab^2-b^3\)
Ta có \(\left(a^3-3ab^2\right)^2\) =\(a^6-6a^4b^2+9a^2b^4=25\)
\(\left(b^3-3a^2b\right)^2=b^6-6a^2b^4+9a^4b^2=100\)
\(=>\left(a^3-3a^2b\right)^2-\left(b^3-3a^2b\right)^2=a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=125\)
\(< =>a^6+3a^4b^2=3a^2b^4+b^6=125\)
\(< =>\left(a^2+b^2\right)^3=125\)
\(=>a^2+b^2=5\)
Ta có:
\(\left(a^3+3ab^2\right)^2=a^6+6a^4b^2+9a^2b^4=196\)
\(\left(b^3+3a^2b\right)^2=b^6+6a^2b^4+9a^4b^2=169\)
Lại có:
\(\left(a^3+3ab^2\right)^2-\left(b^3+3a^2b\right)^2=27\)
\(\Leftrightarrow a^6+6a^4b^2+9ab^4-b^6-6a^2b^4-9a^4b^2=27\)
\(\Leftrightarrow a^6-3a^4b^2+3a^2b^4-b^6=27\)
\(\Leftrightarrow\left(a^2-b^2\right)^3=27\)
\(\Leftrightarrow a^2-b^2=\sqrt[3]{27}=3\)
\(a^3-3ab^2=46\)\(\Rightarrow\left(a^3-3ab^2\right)=46^2\)\(\Rightarrow a^6-6a^4b^2+9a^2b^4=2116\)
\(b^3-3a^2b=9\Rightarrow\left(b^3-3a^2b\right)^2=9^2\Rightarrow b^6-6a^2b^4+9a^4b^2=81\)
\(\Rightarrow a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=2197\)
\(\Rightarrow a^6+3a^4b^2+3a^2b^4+b^6=2197\)
\(\Rightarrow\left(a^2+b^2\right)^3=2197\)
\(\Rightarrow a^2+b^2=13\)
\(M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+b^2+2ab-2ab\right)+6a^2b^2\left(a+b\right)\)
\(M=a^2+2ab+b^2-3ab+3ab-6a^2b^2+6a^2b^2\)
\(M=\left(a+b\right)^2=1\)
Ta có : \(\left(a^2+b^2\right)^3=a^6+3a^4b^2+3a^2b^4+b^6\)
\(=\left(a^6-6a^4b^2+9a^2b^4\right)+\left(b^6-6a^2b^4+9a^4b^2\right)\)
\(=\left(a^3-3ab^2\right)^2+\left(b^3-3a^2b\right)^2\)
\(=5^2+10^2\)
\(=125\)
\(\Rightarrow S^3=125\)
\(\Rightarrow S=5\)
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 ^^
\(\hept{\begin{cases}\left(a^3-3ab^2\right)^2=25\\\left(b^3-3a^2b\right)^2=100\end{cases}}\Leftrightarrow\hept{\begin{cases}a^6-6a^4b^2+9a^2b^4=25\\b^6-6a^2b^4+9a^4b^2=100\end{cases}}\)
Cộng 2 đẳng thức lại ta được:
\(a^6+3a^4b^2+3a^2b^4+b^6=125\Leftrightarrow\left(a^2+b^2\right)^3=125\Leftrightarrow a^2+b^2=5\)
\(\Rightarrow P=2018\left(a^2+b^2\right)=2018.5=...\)
Ta có : \(a^3-3ab^2=5\)
\(\Rightarrow\left(a^3-3ab^2\right)^2=a^6-6a^4b^2+9a^2b^4=25\)
Và \(b^3-3a^2b=10\)
\(\Rightarrow\left(b^3-3a^2b\right)^2=b^6-6a^4b^2+9a^4b^2=100\)
Suy ra : \(a^6++3a^2b^4+3a^4b^2+b^6=125\)
Hoặc : \(\left(a^2+b^2\right)^3=125\Rightarrow a^2+b^2=5\)
Do đó : \(P=2018a^2+2018b^2=2018\left(a^2+b^2\right)=2018.5=10090\)