登录
19. (9分)小丽根据学习“数与式”积累的经验,想通过“由特殊到一般”的方法探究下面二次根式的运算规律,下面是小丽的探究过程,请补充完整:
(1)具体运算,发现规律.
等式1: $\sqrt{1+\frac{1}{3}}=2\sqrt{\frac{1}{3}}$;等式2: $\sqrt{2+\frac{1}{4}}=3\sqrt{\frac{1}{4}}$;
等式3: $\sqrt{3+\frac{1}{5}}=4\sqrt{\frac{1}{5}}$;等式4:
(2)观察、归纳,得出猜想.
n为正整数,猜想等式n可表示为
(3)应用运算规律.
化简:
(1)具体运算,发现规律.
等式1: $\sqrt{1+\frac{1}{3}}=2\sqrt{\frac{1}{3}}$;等式2: $\sqrt{2+\frac{1}{4}}=3\sqrt{\frac{1}{4}}$;
等式3: $\sqrt{3+\frac{1}{5}}=4\sqrt{\frac{1}{5}}$;等式4:
$\sqrt{4+\dfrac{1}{6}} = 5\sqrt{\dfrac{1}{6}}$
.(2)观察、归纳,得出猜想.
n为正整数,猜想等式n可表示为
$\sqrt{n+\dfrac{1}{n+2}} = (n+1)\sqrt{\dfrac{1}{n+2}}$
,并证明你的猜想.(3)应用运算规律.
化简:
答案
(1) $\sqrt{4+\dfrac{1}{6}} = 5\sqrt{\dfrac{1}{6}}$(2分)
(2) $\sqrt{n+\dfrac{1}{n+2}} = (n+1)\sqrt{\dfrac{1}{n+2}}$(4分)
证明: 左边$= \sqrt{\dfrac{n^2+2n+1}{n+2}} = \sqrt{\dfrac{(n+1)^2}{n+2}} = (n+1)\sqrt{\dfrac{1}{n+2}} =$右边.(6分)
(3) 原式$=100\sqrt{\dfrac{1}{101}} × 200\sqrt{\dfrac{1}{201}} × \sqrt{402} × \sqrt{101}$
$=(100×200)×(\sqrt{\dfrac{1}{101}} × \sqrt{101})×(\sqrt{\dfrac{1}{201}} × \sqrt{201})× \sqrt{2}$
$=20\ 000\sqrt{2} .$(9分)
(2) $\sqrt{n+\dfrac{1}{n+2}} = (n+1)\sqrt{\dfrac{1}{n+2}}$(4分)
证明: 左边$= \sqrt{\dfrac{n^2+2n+1}{n+2}} = \sqrt{\dfrac{(n+1)^2}{n+2}} = (n+1)\sqrt{\dfrac{1}{n+2}} =$右边.(6分)
(3) 原式$=100\sqrt{\dfrac{1}{101}} × 200\sqrt{\dfrac{1}{201}} × \sqrt{402} × \sqrt{101}$
$=(100×200)×(\sqrt{\dfrac{1}{101}} × \sqrt{101})×(\sqrt{\dfrac{1}{201}} × \sqrt{201})× \sqrt{2}$
$=20\ 000\sqrt{2} .$(9分)
