第109页 - 苏科版数学补充习题八年级上下册答案

$解：原式 ={\frac {({\sqrt {5}+1})({\sqrt {5}-1})} {4}}$

$={\frac {{({\sqrt {5}})}^{2}-{1}^{2}} {4}}$

$={\frac {5-1} {4}}$

$=1$

$解：原式 ={\frac {{({\sqrt {3}+1})}^{2}+{({\sqrt {3}-1})}^{2}} {4}}$

$={\frac {(3+1+2{\sqrt {3}})+(3+1-2{\sqrt {3}})} {4}}$

$={\frac {4+2{\sqrt {3}+4-2{\sqrt {3}}}} {4}}$

$=2$

$解：(1) 原式 ={(x+y)}^{2}$

$将x=\sqrt {3}+1，y=\sqrt {3}-1代入，得$

$原式 =\left [ {{({\sqrt {3}+1})+({\sqrt {3}-1})}} \right ]^{2}$

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

$=12$

$(2) 原式 =(x+y)(x-y)$

$将x=\sqrt {3}+1，y=\sqrt {3}-1代入，得$

$原式 =\left [ {{({\sqrt {3}+1})+({\sqrt {3}}-1)}} \right ]×\left [ {{({\sqrt {3}+1})-({\sqrt {3}-1})}} \right ]$

$=2{\sqrt {3}}×2$

$=4{\sqrt {3}}$