Đặt \(A=a^2+ab+b^2-3a-3b+1989\)

\(\Rightarrow4A=4a^2+4ab+4b^2-12a-12b+7956\)

\(\Rightarrow4A=\left(4a^2+4ab+b^2\right)-\left(12a+6b\right)+9+\left(3b^2-6b+3\right)+7944\)

\(\Rightarrow4A=\left(2a+b\right)^2-6\left(2a+b\right)+9+3\left(b^2-2b+1\right)+7944\)

\(\Rightarrow4A=\left(2a+b-3\right)^2+3\left(b-1\right)^2+7944\)

\(\Rightarrow A=\frac{\left(2a+b-3\right)^2}{4}+\frac{3\left(b-1\right)^2}{4}+1986\)

Vì \(\frac{\left(2a+b-3\right)^2}{4}\ge0\)\(\forall a,b\);  \(\frac{3\left(b-1\right)^2}{4}\ge0\)\(\forall b\)

\(\Rightarrow\frac{\left(2a+b-3\right)^2}{4}+\frac{3\left(b-1\right)^2}{4}\ge0\)\(\forall a,b\)

\(\Rightarrow\frac{\left(2a+b-3\right)^2}{4}+\frac{3\left(b-1\right)^2}{4}+1986\ge1986\)\(\forall a,b\)

\(\Rightarrow A\ge1986\)\(\forall a,b\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{\left(2a+b-3\right)^2}{4}=0\\\frac{3\left(b-1\right)^2}{4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(2a+b-3\right)^2=0\\3\left(b-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2a+b-3=0\\b-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}2a+b-3=0\\b=1\end{cases}}\Leftrightarrow\hept{\begin{cases}2a+1-3=0\\b=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=1\end{cases}}\Leftrightarrow a=b=1\)

Vậy.....

SO4 kia là nhóm nguyên tử SO4 nhé

Bài làm 

\(2AL+3SO_4\rightarrow AL_2\left(SO_4\right)_3\)

Học tốt 

Ta  có : 99ⁿ  + 99^{n + 1} = 99ⁿ + 99ⁿ.99 = 99ⁿ(99 + 1) = 99ⁿ . 100\(⋮\)100 (đpcm)