第5页 - 补充习题江苏八年级数学人教版人民教育出版社

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解$：(1)$原式$=3 \sqrt {3}+\frac {\sqrt {3}}{3} -(2 \sqrt {3}-\frac {\sqrt {5}}{5}+3 \sqrt {5}) $

$=\frac {10 \sqrt {3}}{3}-2 \sqrt {3}+\frac {14 \sqrt {5}}{5} $

$=\frac {10 \sqrt {3}}{3}-2 \sqrt {3}-\frac {14 \sqrt {5}}{5} $

$=\frac {4 \sqrt {3}}{3}-\frac {14 \sqrt {5}}{5}$

$(2)2 \sqrt {125}-\sqrt {28}+\sqrt {\frac {1}{20}}-\frac {1}{3} \sqrt {175} $

$=10 \sqrt {5}-2 \sqrt {7}+\frac {\sqrt {5}}{10}-\frac {5 \sqrt {7}}{3} $

$=\frac {101 \sqrt {5}}{10}-\frac {11 \sqrt {7}}{3}$

$\text { (3)原式 }=2 \sqrt{a-b}+(a-b) \sqrt{a-b}-a \sqrt{a-b} $

$=(2-b) \sqrt {a-b}$

解$：\frac {2}{5} \sqrt {25\ \mathrm {a}}+9 \sqrt {\frac {a}{3}}-2\ \mathrm {a}^2 \sqrt {\frac {1}{a^3}}=18 $

$\frac {2}{5} \sqrt {5^2 ·a}+9 \sqrt {\frac {3\ \mathrm {a}}{3^2}}-2\ \mathrm {a}^2\sqrt {\frac {a}{(a^2)^2}} =18 $

$\frac {2}{5} ×5 \sqrt {a}+9 ×\frac {1}{3} \sqrt {3\ \mathrm {a}}-2\ \mathrm {a}^2 ·\frac {1}{a^2} \sqrt {a}=18 $

$2 \sqrt {a}+3 \sqrt {3\ \mathrm {a}}-2 \sqrt {a}=18 $

$3 \sqrt {3\ \mathrm {a}}=18 $

$\sqrt {3\ \mathrm {a}}=6 $

$3\ \mathrm {a}=6^2 $

$a=12$

解:∵$x+\frac {1}{x}=\sqrt {8} $

∴$(x+\frac {1}{x})^2=x^2+2+\frac {1}{x^2}=8 $

∴$x^2+\frac {1}{x^2}=6 $

∴$x^2+\frac {1}{x^2}-2=4 $

$\text { 即 }(x-\frac{1}{x})^{2}=4 $

∴$x-\frac{1}{x}=2 \text { 或 }-2$

解:∵$x=2-\sqrt {3}， y=2+\sqrt {3}， $

∴$x^2+x y+y^2 $

$=x^2+2 x y+y^2-x y $

$=(x+y)^2-x y $

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

$-(2-\sqrt {3})(2+\sqrt {3}) $

$=16-4+3 $

$=15 .$

解$：(1)x^2=24 $

$x= \pm \sqrt {24} $

$x_1=2 \sqrt {6}， x_2=-2 \sqrt {6}$

$\text { (2) } 3 \sqrt{2} x=-\sqrt{8} $

$x=-\sqrt {8} ·\frac {1}{3 \sqrt {2}} $

$x=-\frac {2}{3}$

解$：(1)$原式$=5-3=2；$

$(2)$原式$=16+3+8\sqrt 3-16-3+8\sqrt 3=16\sqrt 3；$

$(3)$原式$=\frac 12\sqrt 3-2\sqrt 6-2\sqrt 3=-\frac 32\sqrt 3-2\sqrt 6；$

$(4)$原式$=(1+x)^2-x=1+x+x^2$

$\text { 解: } $∵${x}=\sqrt {5}-2， $

∴$x^2=(\sqrt {5}-2)^2=5-4 \sqrt {5}+4=9-4 \sqrt {5}， $

∴$(9+4 \sqrt {5}) x^2-(\sqrt {5}+2) {x}+4$

$=(9+4 \sqrt {5})(9-4 \sqrt {5})-( \sqrt {5}+2)(\sqrt {5}-2)+4 $

$=81-80-(5-4)+4 $

$=1-1+4 $

$=4 .$

解:∵$a-b=\sqrt {3}+\sqrt {2}， b-c=\sqrt {3}-\sqrt {2}， $

∴$a-c=2 \sqrt {3}， $

∴$a^2+b^2+c^2-a b-b c-a c $

$=\frac {1}{2}[(a-b)^2+(b-c)^2+(a-c)^2] $

$=\frac {1}{2}[(\sqrt {3}+\sqrt {2})^2+(\sqrt {3}-\sqrt {2})^2. .+(2 \sqrt {3})^2] $

$=\frac {1}{2}(5+2 \sqrt {6}+5-2 \sqrt {6}+12) $

$=\frac {1}{2} ×22 $

$=11 .$