第18页

信息发布者：
$4\sqrt{6}$
​$ -\frac {2}{3}$​
$-2a$
解：
​$ $​原式​$=(1-\frac {1}{x+2})÷\frac {x^2-1}{x^2+4x+4}$​
​$ =\frac {x+2-1}{x+2}÷\frac {(x-1)(x+1)}{(x+2)^2}$​
​$ =\frac {x+1}{x+2}·\frac {(x+2)^2}{(x-1)(x+1)}$​
​$ =\frac {x+2}{x-1}$​
​$ $​当​$x=\sqrt {3}+1$​时，
​$ $​原式​$=\frac {\sqrt {3}+1+2}{\sqrt {3}+1-1}=\frac {\sqrt {3}+3}{\sqrt {3}}=1+\sqrt {3}$​
解：
原式$=(a-b)^2+2a(a+b)+(a+2b)(a-2b)$
$=a^2-2ab+b^2+2a^2+2ab+a^2-4b^2$
$=4a^2-3b^2$
当$a=\sqrt{2},b=\sqrt{3}$时，
原式$=4×(\sqrt{2})^2-3×(\sqrt{3})^2=8-9=-1$
D
$-25$
解：
$\because 6+2\sqrt{5}=(1+\sqrt{5})^2，$
$\therefore$原式$=(1+\sqrt{5})^2x^2+(1+\sqrt{5})x+\sqrt{5}$
$=[(1+\sqrt{5})x]^2+(1+\sqrt{5})x+\sqrt{5}$
当$x=1-\sqrt{5}$时，
$(1+\sqrt{5})x=(1+\sqrt{5})(1-\sqrt{5})=1-5=-4$
则原式$=(-4)^2+(-4)+\sqrt{5}=16-4+\sqrt{5}=12+\sqrt{5}$
解： ∵​$x=\frac {1}{5-2\sqrt {6}}=\frac {5+2\sqrt {6}}{(5-2\sqrt {6})(5+2\sqrt {6})}=5+2\sqrt {6}，$​
​$ y=\frac {1}{5+2\sqrt {6}}=\frac {5-2\sqrt {6}}{(5+2\sqrt {6})(5-2\sqrt {6})}=5-2\sqrt {6}，$​
∴​$x+y=10，$​​$xy=(5+2\sqrt {6})(5-2\sqrt {6})=25-24=1$​
​$ $​原式​$=\frac {x^2+y^2}{xy}-4=\frac {(x+y)^2-2xy}{xy}-4$​
​$ =\frac {10^2-2×1}{1}-4=100-2-4=94$​
解：
∵​$x+\frac {1}{x}=\sqrt {7}，$​∴​$x≠0$​
​$ \frac {x^4+x^2+1}{x^2}=x^2+\frac {1}{x^2}+1=(x+\frac {1}{x})^2-1$​
​$ =(\sqrt {7})^2-1=7-1=6$​
∴原式​$=\frac {1}{6}$​
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