$C=\underbrace{11...1}_{\text{n số 1}}\underbrace{55...5}_{\text{n số 5}}+1 \\
=\underbrace{11...1}_{\text{n số 1}}\underbrace{00...0}_{\text{n số 0}}+\underbrace{55...5}_{\text{n số 5}}+1 \\
=\underbrace{11...1}_{\text{n số 1}}.1\underbrace{00...0}_{\text{n số 0}}+\underbrace{55...5}_{\text{n số 5}}+1 \\
=\dfrac{\underbrace{99...9}_{\text{n số 9}}}9.1\underbrace{00...0}_{\text{n số 0}}+\dfrac{\underbrace{99...9}_{\text{n số 9}}}9.5+1 \\
=\dfrac{1\underbrace{00...0}_{\text{n số 0}}-1}9.1\underbrace{00...0}_{\text{n số 0}}+\dfrac{1\underbrace{00...0}_{\text{n số 0}}-1}9.5+1 \\
=\dfrac{10^n-1}9.10^n+\dfrac{10^n-1}9.5+\dfrac99 \\
=\dfrac{(10^n-1).10^n+(10^n-1).5+9}9 \\
=\dfrac{10^{2n}-10^n+5.10^n-5+9}9 \\
=\dfrac{{(10^n)}^2+4.10^n+4}9 \\
=\dfrac{(10^n+2)^2}9 \\
={\bigg(\dfrac{10^n+2}3\bigg)}^2
$
Chắc đề là : $A=\underbrace{11...1}_{\text{2n số 1}}-\underbrace{88...8}_{\text{n số 8}}+1 \\
=\dfrac{10^{2n}-1}9-\dfrac{10^n-1}9.8+\dfrac99 \\
=\dfrac{10^{2n}-1-8(10^n-1)+9}9 \\
=\dfrac{10^{2n}-1-8.10^n+8+9}9 \\
=\dfrac{{(10^n)}^2-8.10^n+16}9 \\
=\dfrac{(10^n-4)^2}9 \\
={\bigg(\dfrac{10^n-4}3\bigg)}^2$
$B=\underbrace{11...1}_{\text{2n số 1}}+\underbrace{11...1}_{\text{n+1 số 1}}+\underbrace{66...6}_{\text{n số 6}}+8 \\
=\dfrac{10^{2n}-1}9+\dfrac{10^{n+1}-1}9+\dfrac{10^n-1}9.6+\dfrac{72}9 \\
=\dfrac{10^{2n}-1+10^{n+1}-1+6(10^n-1)+72}9 \\
=\dfrac{10^{2n}+10^{n+1}+6.10^n-6+70}9 \\
=\dfrac{10^{2n}+10^n(10+6)+64}9 \\
=\dfrac{(10^n)^2+16.10^n+64}9 \\
=\dfrac{(10^n+8)^2}9 \\
={\bigg(\dfrac{10^n+8}3\bigg)}^2$