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19. 观察下列各式:①$\sqrt{1+\frac{1}{3}}=2\sqrt{\frac{1}{3}}$,②$\sqrt{2+\frac{1}{4}}=3\sqrt{\frac{1}{4}}$,③$\sqrt{3+\frac{1}{5}}=4\sqrt{\frac{1}{5}},\dots$
(1)请观察规律,并写出第④个等式:________;
(2)请用含$n(n≥1)$的式子写出你猜想的规律:________;
(3)请证明(2)中的结论.
(1)请观察规律,并写出第④个等式:________;
(2)请用含$n(n≥1)$的式子写出你猜想的规律:________;
(3)请证明(2)中的结论.
答案
19.(1)$\sqrt{4+\frac{1}{6}}=5\sqrt{\frac{1}{6}}$
(2)$\sqrt{n+\frac{1}{n+2}}=(n+1)\sqrt{\frac{1}{n+2}}$
(3)证明:$\sqrt{n+\frac{1}{n+2}}=\sqrt{\frac{n^2+2n}{n+2}+\frac{1}{n+2}}=\sqrt{\frac{n^2+2n+1}{n+2}}=\sqrt{\frac{(n+1)^2}{n+2}}$.
$\because n≥1$,$\therefore$原式$=(n+1)\sqrt{\frac{1}{n+2}}$.
(2)$\sqrt{n+\frac{1}{n+2}}=(n+1)\sqrt{\frac{1}{n+2}}$
(3)证明:$\sqrt{n+\frac{1}{n+2}}=\sqrt{\frac{n^2+2n}{n+2}+\frac{1}{n+2}}=\sqrt{\frac{n^2+2n+1}{n+2}}=\sqrt{\frac{(n+1)^2}{n+2}}$.
$\because n≥1$,$\therefore$原式$=(n+1)\sqrt{\frac{1}{n+2}}$.
