chứng minh rằng mỗi số dưới đây là bình phương của một số tự nhiên
a/A= 99….900…025
n n
b/ B= 99…9800…01
n n
c/ C= 44…488…89
n n-1
d/ D= 11…122…25
n n+1
chứng minh rằng mỗi số dưới đây là bình phương của một số tự nhiên a/A= 99….900…025 n n b/ B= 99…9800…01
By Nevaeh
a, A = 99…9 . 10n+210n+2 + 25
= (10n10n – 1).10n+210n+2 + 25
= 102n+2102n+2 – 10n+210n+2 + 25
= (10n+110n+1² – 2.5.10n+110n+1 + 5²
= (10n+110n+1 – 5)²
b, B = 99…9 . 10n+210n+2 + 8.10n+110n+1 + 1
= (10n10n – 1).10n+210n+2 + 8.10n+110n+1 + 1
= 102n+2102n+2 – 10.10n+110n+1 + 8.10n+110n+1 + 1
= 102n+2102n+2 – 2.10n+110n+1 + 1
= (10n+110n+1 – 1)²
c, C = 44…4 . 10n10n + 88…8 . 10 + 9
= 4949.(10n10n – 1).10n10n + 8989.(10n−110n−1 – 1).10 + 9
= 49102n−4910n+8910n−809+949102n−4910n+8910n−809+9
= 49102n+4910n+1949102n+4910n+19
= [13(2.10n+1)]2[13(2.10n+1)]2
d, D = 19.(10n−1).10n+2+29.(10n+1−1).10+519.(10n−1).10n+2+29.(10n+1−1).10+5
= 19.102n+2−19.10n+2+29.10n+2−209+519.102n+2−19.10n+2+29.10n+2−209+5
= 19.102n+2+109.10n+1+25919.102n+2+109.10n+1+259
= [13(10n+1+5)]2
Giải thích các bước giải:
a, A = 99…9 . \({10^{n + 2}}\) + 25
= (\({10^n}\) – 1).\({10^{n + 2}}\) + 25
= \({10^{2n + 2}}\) – \({10^{n + 2}}\) + 25
= (\({10^{n + 1}}\)² – 2.5.\({10^{n + 1}}\) + 5²
= (\({10^{n + 1}}\) – 5)²
b, B = 99…9 . \({10^{n + 2}}\) + 8.\({10^{n + 1}}\) + 1
= (\({10^n}\) – 1).\({10^{n + 2}}\) + 8.\({10^{n + 1}}\) + 1
= \({10^{2n + 2}}\) – 10.\({10^{n + 1}}\) + 8.\({10^{n + 1}}\) + 1
= \({10^{2n + 2}}\) – 2.\({10^{n + 1}}\) + 1
= (\({10^{n + 1}}\) – 1)²
c, C = 44…4 . \({10^n}\) + 88…8 . 10 + 9
= \(\frac{4}{9}\).(\({10^n}\) – 1).\({10^n}\) + \(\frac{8}{9}\).(\({10^{n – 1}}\) – 1).10 + 9
= \(\frac{4}{9}{10^{2n}} – \frac{4}{9}{10^n} + \frac{8}{9}{10^n} – \frac{{80}}{9} + 9\)
= \(\frac{4}{9}{10^{2n}} + \frac{4}{9}{10^n} + \frac{1}{9}\)
= \({{\rm{[}}\frac{1}{3}({2.10^n} + 1){\rm{]}}^2}\)
d, D = \(\frac{1}{9}.({10^n} – 1){.10^{n + 2}} + \frac{2}{9}.({10^{n + 1}} – 1).10 + 5\)
= \(\frac{1}{9}{.10^{2n + 2}} – \frac{1}{9}{.10^{n + 2}} + \frac{2}{9}{.10^{n + 2}} – \frac{{20}}{9} + 5\)
= \(\frac{1}{9}{.10^{2n + 2}} + \frac{{10}}{9}{.10^{n + 1}} + \frac{{25}}{9}\)
= \({{\rm{[}}\frac{1}{3}({10^{n + 1}} + 5){\rm{]}}^2}\)