第141页 - 江苏版实验班提优训练数学八年级上下册答案

$\sqrt{n+\frac{1}{n+2}}=(n+1)\sqrt{\frac{1}{n+2}}$

$ \begin{aligned} 解:(1)原式&=2(a²-3)-a²+\sqrt {2} a+6\ \\ &=2a²-6-a²+\sqrt {2} a+6=a²+\sqrt {2} a.\ \\ \end{aligned}$

$当a=\sqrt {2} -1时，$

$ \begin{aligned} 原式&=(\sqrt{2}-1)²+\sqrt{2}(\sqrt{2}-1) \\ &=3-2\sqrt{2}+2-\sqrt{2}=5-3\sqrt{2}. \\ \end{aligned}$

$(2)∵x=2+\sqrt {3} ，y=2-\sqrt {3} ，\ $

$∴x+y=4，xy=(2+\sqrt{3})(2-\sqrt {3} )=1.\ $

$ \begin{aligned}∴x²+y²-3xy&=x²+2xy+y²-5xy \\ &=(x+y)²-5xy=16-5=11. \\ \end{aligned}$

$解:∵x=\frac{\sqrt{5}-1}{2}，$

$∴2x=\sqrt {5} -1.$

$∴2x+1=\sqrt {5} .\ $

$两边平方，得(2x+1)²=5，\ $

$即4x²+4x+1=5，$

$∴4x²+4x=4，$

$∴x²+x=1，\ $

$∴x^{3} +2x²=x^{3} +x²+x²=x(x²+x)+x²$

$=x+x²=1.$

（更多请点击查看作业精灵详解）

$ \begin{aligned}解:原式&=2\sqrt {3}×3\sqrt {2}-2\sqrt {3}×1+\sqrt {6} ×3\sqrt {2} -\sqrt {6}×1 \\ &=6\sqrt {6}-2\sqrt {3}+6\sqrt {3}-\sqrt {6} \\ &=5\sqrt {6} +4\sqrt {3} \\ \end{aligned}$

$ \begin{aligned}解:原式&=(\frac {5}{4}×\frac {2\sqrt {5} }{5}-4\sqrt {5} )+(5×\frac {\sqrt {5} }{5}-\frac {1}{2}×3\sqrt {5} ) \\ &=(\frac {1}{2}\sqrt {5}-4\sqrt {5} )+(\sqrt {5} -\frac {3}{2}\sqrt {5} ) \\ &=-\frac {7}{2}\sqrt {5}-\frac {1}{2}\sqrt {5} \\ &=-4\sqrt {5} \\ \end{aligned}$

$解：原式=2×1+\sqrt 2-\sqrt 2$

$~~~~~~~~~~~~~~~~~=2$

$解：原式=(\sqrt 3-\sqrt 2){(\sqrt 3-\sqrt 2)(\sqrt 3+\sqrt 2)}^{2022}$

$~~~~~~~~~~~~~~~~~=(\sqrt 3-\sqrt 2)(3-2)^{2022}$

$~~~~~~~~~~~~~~~~~=(\sqrt 3-\sqrt 2)$