\(\hept{\begin{cases}\\\end{cases}\hept{\begin{cases}\\\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}^{ }^2_{ }\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\end{cases}}}\)

`Answer:`

a) \(\frac{5}{18}x^2y.18x^3y^2\)

\(=\left(\frac{5}{18}.18\right).\left(x^2.x^3\right).\left(y.y^2\right)\)

\(=5x^5y^3\)

Bậc: `8`

b) \(\frac{2}{9}xy^2.\left(-36x^2y^3\right)\)

\(=\left(\frac{2}{9}.-36\right).\left(x.x^2\right).\left(y^2.y^3\right)\)

\(=-8x^3y^5\)

Bậc: `8`

`Answer:`

\(a^2+2b^2-2ab+2a-4b+2=0\)

\(\Leftrightarrow a^2+b^2+b^2-2ab+2a-2b-2b+1+1=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(2a-2b\right)+1+\left(b^2-2b+1\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+2.\left(a-b\right)+1+\left(b-1\right)^2=0\)

\(\Leftrightarrow\left(a-b+1\right)^2+\left(b-1\right)^2=0\)

Mà \(\hept{\begin{cases}\left(a-b+1\right)^2\ge0\forall a,b\\\left(b-1\right)^2\ge0\forall b\end{cases}}\Rightarrow\left(a-b+1\right)^2+\left(b-1\right)^2\ge0\forall a,b\)

Ta có: \(\left(a-b+1\right)^2+\left(b-1\right)^2=0\)

\(\hept{\begin{cases}a-b+1=0\\b-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a-1+1=0\\b=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=0\\b=1\end{cases}}\)

Ta có: \(\left(a+b\right)^3=\overline{ab^2}\) là số chính phương nên a + b là số chính phương

Đặt \(a+b=x^2\) với \(x\inℕ^∗\)

\(\Rightarrow\overline{ab^2}=x^6\)

\(\Rightarrow x^3=\overline{ab}< 100\) và \(\overline{ab}>9\)

\(\Rightarrow9< \overline{ab}< 100\)

\(\Rightarrow9< x^3< 100\)

\(\Rightarrow2< x< 5\)

\(\Rightarrow x=3\) hoặc \(x=4\)

Với \(x=3\Rightarrow\overline{ab^2}=\left(a+b\right)^3=x^6=3^6=729=27^2=\left(2+7\right)^3=\left(TM\right)\)

Với \(x=4\Rightarrow\overline{ab}^2=\left(a+b\right)^3=x^6=4^6=4096=64^2\ne\left(6+4\right)^3\left(KTM\right)\)

Vậy số cần tìm là: 27