Question

tinh \(\sqrt{1+99...9^2+0,99...99^2}\)(n c/s 9)

Answer 1

\(99...9=10^n-1\)(n chữ số 9)

\(0,99...9=1-\dfrac{1}{10^n}\)(n chữ số 9)

\(\sqrt{1+99...9^2+0.99...99^2}\\ =\sqrt{1+\left(10^n-1\right)^2+\left(1-\dfrac{1}{10^n}\right)^2}\\ =\sqrt{1+10^{2n}+1-2.10^n+1+\dfrac{1}{10^{2n}}-\dfrac{2}{10^n}}\\ =\sqrt{3+10^{2n}-2.10^n+\dfrac{1}{10^{2n}}-\dfrac{2}{10^n}}\\ =\sqrt{\dfrac{3.10^{2n}+10^{4n}-2.10^{3n}+1-2.10^n}{10^{2n}}}\\ =\sqrt{\dfrac{\left(10^{2n}-10^n+1\right)^2}{10^{2n}}}=\dfrac{10^{2n}-10^n+1}{10^n}\\ =10^n-1+\dfrac{1}{10^n}=99...9+1-0,99...9=99...9,00...1\)

(n chữ số 9,n-1 chữ số 0)