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8. 计算：$|1-\sqrt{2}|-(\sqrt{3})^{0}+\dfrac{4}{\sqrt{3}-1}-(\dfrac{\sqrt{2}}{2})^{-1}$。
答案
$|1 - \sqrt{2}| = \sqrt{2} - 1$（因为 $\sqrt{2} ＞ 1$），
$(\sqrt{3})^{0} = 1$（任何非零数的0次方都是1），
$\frac{4}{\sqrt{3} - 1} = \frac{4(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{4(\sqrt{3} + 1)}{3 - 1} = 2(\sqrt{3} + 1)=2\sqrt{3} + 2$（分母有理化），
$(\frac{\sqrt{2}}{2})^{-1} = \sqrt{2} × \frac{2}{2} =\sqrt{2} ×1= \sqrt{2}$（负指数表示取倒数后的正值），
原式$= (\sqrt{2} - 1) - 1 + (2\sqrt{3} + 2) - \sqrt{2}$
$= \sqrt{2} - 1 - 1 + 2\sqrt{3} + 2 - \sqrt{2}$
$= 2\sqrt{3}$。
$(\sqrt{3})^{0} = 1$（任何非零数的0次方都是1），
$\frac{4}{\sqrt{3} - 1} = \frac{4(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{4(\sqrt{3} + 1)}{3 - 1} = 2(\sqrt{3} + 1)=2\sqrt{3} + 2$（分母有理化），
$(\frac{\sqrt{2}}{2})^{-1} = \sqrt{2} × \frac{2}{2} =\sqrt{2} ×1= \sqrt{2}$（负指数表示取倒数后的正值），
原式$= (\sqrt{2} - 1) - 1 + (2\sqrt{3} + 2) - \sqrt{2}$
$= \sqrt{2} - 1 - 1 + 2\sqrt{3} + 2 - \sqrt{2}$
$= 2\sqrt{3}$。
9. 比较大小：$\sqrt{18}-\sqrt{17}$
$＞$
$\sqrt{19}-\sqrt{18}$.(填“$＞$”“$＜$”或“$=$”)答案
$＞$
解析
因为$\sqrt{18}-\sqrt{17}=\frac{(\sqrt{18}-\sqrt{17})(\sqrt{18}+\sqrt{17})}{\sqrt{18}+\sqrt{17}}=\frac{18 - 17}{\sqrt{18}+\sqrt{17}}=\frac{1}{\sqrt{18}+\sqrt{17}}$，
同理$\sqrt{19}-\sqrt{18}=\frac{(\sqrt{19}-\sqrt{18})(\sqrt{19}+\sqrt{18})}{\sqrt{19}+\sqrt{18}}=\frac{19 - 18}{\sqrt{19}+\sqrt{18}}=\frac{1}{\sqrt{19}+\sqrt{18}}$，
又因为$\sqrt{18}+\sqrt{17}\lt\sqrt{19}+\sqrt{18}$，
根据分子相同，分母越大分数越小，可得$\frac{1}{\sqrt{18}+\sqrt{17}}\gt\frac{1}{\sqrt{19}+\sqrt{18}}$，
即$\sqrt{18}-\sqrt{17}\gt\sqrt{19}-\sqrt{18}$。
同理$\sqrt{19}-\sqrt{18}=\frac{(\sqrt{19}-\sqrt{18})(\sqrt{19}+\sqrt{18})}{\sqrt{19}+\sqrt{18}}=\frac{19 - 18}{\sqrt{19}+\sqrt{18}}=\frac{1}{\sqrt{19}+\sqrt{18}}$，
又因为$\sqrt{18}+\sqrt{17}\lt\sqrt{19}+\sqrt{18}$，
根据分子相同，分母越大分数越小，可得$\frac{1}{\sqrt{18}+\sqrt{17}}\gt\frac{1}{\sqrt{19}+\sqrt{18}}$，
即$\sqrt{18}-\sqrt{17}\gt\sqrt{19}-\sqrt{18}$。
10. 已知 $a= \dfrac{1}{\sqrt{7}+2\sqrt{2}}$,$b= \dfrac{1}{2\sqrt{2}-\sqrt{7}}$,求下列代数式的值：
(1) $a^{2}b+ab^{2}$；
(2) $a^{2}+b^{2}$；
(3) $\dfrac{a-b}{a-2\sqrt{ab}+b}$。
(1) $a^{2}b+ab^{2}$；
(2) $a^{2}+b^{2}$；
(3) $\dfrac{a-b}{a-2\sqrt{ab}+b}$。
答案
(1) 先化简 $a$ 和 $b$：
$a = \dfrac{1}{\sqrt{7} + 2\sqrt{2}} = \dfrac{2\sqrt{2} - \sqrt{7}}{(2\sqrt{2})^2 - (\sqrt{7})^2} = 2\sqrt{2} - \sqrt{7}$，
$b = \dfrac{1}{2\sqrt{2} - \sqrt{7}} = \dfrac{2\sqrt{2} + \sqrt{7}}{(2\sqrt{2})^2 - (\sqrt{7})^2} = 2\sqrt{2} + \sqrt{7}$。
则 $a + b = 4\sqrt{2}$，$ab = (2\sqrt{2})^2 - (\sqrt{7})^2 = 1$。
$a^2b + ab^2 = ab(a + b) = 1 × 4\sqrt{2} = 4\sqrt{2}$。
(2) $a^2 + b^2 = (a + b)^2 - 2ab = (4\sqrt{2})^2 - 2 × 1 = 32 - 2 = 30$。
(3) $a - b = (2\sqrt{2} - \sqrt{7}) - (2\sqrt{2} + \sqrt{7}) = -2\sqrt{7}$，
分母 $a - 2\sqrt{ab} + b = (a + b) - 2\sqrt{ab} = 4\sqrt{2} - 2$。
原式 $= \dfrac{-2\sqrt{7}}{4\sqrt{2} - 2} = \dfrac{-2\sqrt{7}}{2(2\sqrt{2} - 1)} = \dfrac{-\sqrt{7}}{2\sqrt{2} - 1} = \dfrac{-\sqrt{7}(2\sqrt{2} + 1)}{(2\sqrt{2})^2 - 1^2} = \dfrac{-2\sqrt{14} - \sqrt{7}}{7}$。
(1) $4\sqrt{2}$
(2) $30$
(3) $\dfrac{-2\sqrt{14} - \sqrt{7}}{7}$
$a = \dfrac{1}{\sqrt{7} + 2\sqrt{2}} = \dfrac{2\sqrt{2} - \sqrt{7}}{(2\sqrt{2})^2 - (\sqrt{7})^2} = 2\sqrt{2} - \sqrt{7}$，
$b = \dfrac{1}{2\sqrt{2} - \sqrt{7}} = \dfrac{2\sqrt{2} + \sqrt{7}}{(2\sqrt{2})^2 - (\sqrt{7})^2} = 2\sqrt{2} + \sqrt{7}$。
则 $a + b = 4\sqrt{2}$，$ab = (2\sqrt{2})^2 - (\sqrt{7})^2 = 1$。
$a^2b + ab^2 = ab(a + b) = 1 × 4\sqrt{2} = 4\sqrt{2}$。
(2) $a^2 + b^2 = (a + b)^2 - 2ab = (4\sqrt{2})^2 - 2 × 1 = 32 - 2 = 30$。
(3) $a - b = (2\sqrt{2} - \sqrt{7}) - (2\sqrt{2} + \sqrt{7}) = -2\sqrt{7}$，
分母 $a - 2\sqrt{ab} + b = (a + b) - 2\sqrt{ab} = 4\sqrt{2} - 2$。
原式 $= \dfrac{-2\sqrt{7}}{4\sqrt{2} - 2} = \dfrac{-2\sqrt{7}}{2(2\sqrt{2} - 1)} = \dfrac{-\sqrt{7}}{2\sqrt{2} - 1} = \dfrac{-\sqrt{7}(2\sqrt{2} + 1)}{(2\sqrt{2})^2 - 1^2} = \dfrac{-2\sqrt{14} - \sqrt{7}}{7}$。
(1) $4\sqrt{2}$
(2) $30$
(3) $\dfrac{-2\sqrt{14} - \sqrt{7}}{7}$
11. 小明在解决问题“已知 $a= \dfrac{1}{2+\sqrt{3}}$,求 $2a^{2}-8a+1$ 的值”时,他是这样解答的：
解：$\because a= \dfrac{1}{2+\sqrt{3}}= \dfrac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}= 2-\sqrt{3}$,
$\therefore a-2= -\sqrt{3}$,
$\therefore (a-2)^{2}= 3$,即 $a^{2}-4a+4= 3$,
$\therefore a^{2}-4a= -1$,
$\therefore 2a^{2}-8a+1= 2(a^{2}-4a)+1= 2× (-1)+1= -1$。
请你根据小明的解题过程,解决下列问题：
(1) 化简：$\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{5}+\sqrt{3}}+\dfrac{1}{\sqrt{7}+\sqrt{5}}+…+\dfrac{1}{\sqrt{121}+\sqrt{119}}$；
(2) 若 $a= \dfrac{1}{\sqrt{2}-1}$,
①求 $4a^{2}-8a+1$ 的值；
②直接写出代数式的值：
$a^{3}-3a^{2}+a+1= $
$2a^{2}-5a+\dfrac{1}{a}+2= $
解：$\because a= \dfrac{1}{2+\sqrt{3}}= \dfrac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}= 2-\sqrt{3}$,
$\therefore a-2= -\sqrt{3}$,
$\therefore (a-2)^{2}= 3$,即 $a^{2}-4a+4= 3$,
$\therefore a^{2}-4a= -1$,
$\therefore 2a^{2}-8a+1= 2(a^{2}-4a)+1= 2× (-1)+1= -1$。
请你根据小明的解题过程,解决下列问题：
(1) 化简：$\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{5}+\sqrt{3}}+\dfrac{1}{\sqrt{7}+\sqrt{5}}+…+\dfrac{1}{\sqrt{121}+\sqrt{119}}$；
(2) 若 $a= \dfrac{1}{\sqrt{2}-1}$,
①求 $4a^{2}-8a+1$ 的值；
②直接写出代数式的值：
$a^{3}-3a^{2}+a+1= $
0
,$2a^{2}-5a+\dfrac{1}{a}+2= $
2
。答案
(1)
$\begin{aligned}&\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{5}+\sqrt{3}}+\dfrac{1}{\sqrt{7}+\sqrt{5}}+\cdots+\dfrac{1}{\sqrt{121}+\sqrt{119}}\\=&\dfrac{\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)}+\dfrac{\sqrt{5}-\sqrt{3}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}+\dfrac{\sqrt{7}-\sqrt{5}}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}+\cdots+\dfrac{\sqrt{121}-\sqrt{119}}{(\sqrt{121}+\sqrt{119})(\sqrt{121}-\sqrt{119})}\\=&\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+\dfrac{\sqrt{7}-\sqrt{5}}{2}+\cdots+\dfrac{\sqrt{121}-\sqrt{119}}{2}\\=&\dfrac{1}{2}×( \sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\cdots+\sqrt{121}-\sqrt{119})\\=&\dfrac{1}{2}×(11 - 1)\\=&5\end{aligned}$
(2)
因为$a = \dfrac{1}{\sqrt{2}-1}=\dfrac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\sqrt{2}+1$。
①
$\begin{aligned}&4a^{2}-8a + 1\\=&4(a^{2}-2a)+1\\=&4×[(a^{2}-2a + 1)-1]+1\\=&4×[(a - 1)^{2}-1]+1\\=&4×[(\sqrt{2}+1-1)^{2}-1]+1\\=&4×(2 - 1)+1\\=&5\end{aligned}$
②
因为$a=\sqrt{2}+1$，则$a - 1=\sqrt{2}$，$(a - 1)^{2}=2$，$a^{2}-2a+1 = 2$，$a^{2}-2a=1$，$a^{2}=2a + 1$。
$\begin{aligned}&a^{3}-3a^{2}+a + 1\\=&a(a^{2}-3a + 1)+1\\=&a(2a + 1-3a + 1)+1\\=&a(-a + 2)+1\\=&-a^{2}+2a + 1\\=&-(2a + 1)+2a + 1\\=&0\end{aligned}$
$\begin{aligned}&2a^{2}-5a+\dfrac{1}{a}+2\\=&2(2a + 1)-5a+\dfrac{1}{\sqrt{2}+1}+2\\=&4a+2-5a+\sqrt{2}-1 + 2\\=&-a+\sqrt{2}+3\\=&-(\sqrt{2}+1)+\sqrt{2}+3\\=&2\end{aligned}$
综上，答案依次为：(1)$5$；(2)①$5$；②$0$；$2$。
$\begin{aligned}&\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{5}+\sqrt{3}}+\dfrac{1}{\sqrt{7}+\sqrt{5}}+\cdots+\dfrac{1}{\sqrt{121}+\sqrt{119}}\\=&\dfrac{\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)}+\dfrac{\sqrt{5}-\sqrt{3}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}+\dfrac{\sqrt{7}-\sqrt{5}}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}+\cdots+\dfrac{\sqrt{121}-\sqrt{119}}{(\sqrt{121}+\sqrt{119})(\sqrt{121}-\sqrt{119})}\\=&\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+\dfrac{\sqrt{7}-\sqrt{5}}{2}+\cdots+\dfrac{\sqrt{121}-\sqrt{119}}{2}\\=&\dfrac{1}{2}×( \sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\cdots+\sqrt{121}-\sqrt{119})\\=&\dfrac{1}{2}×(11 - 1)\\=&5\end{aligned}$
(2)
因为$a = \dfrac{1}{\sqrt{2}-1}=\dfrac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\sqrt{2}+1$。
①
$\begin{aligned}&4a^{2}-8a + 1\\=&4(a^{2}-2a)+1\\=&4×[(a^{2}-2a + 1)-1]+1\\=&4×[(a - 1)^{2}-1]+1\\=&4×[(\sqrt{2}+1-1)^{2}-1]+1\\=&4×(2 - 1)+1\\=&5\end{aligned}$
②
因为$a=\sqrt{2}+1$，则$a - 1=\sqrt{2}$，$(a - 1)^{2}=2$，$a^{2}-2a+1 = 2$，$a^{2}-2a=1$，$a^{2}=2a + 1$。
$\begin{aligned}&a^{3}-3a^{2}+a + 1\\=&a(a^{2}-3a + 1)+1\\=&a(2a + 1-3a + 1)+1\\=&a(-a + 2)+1\\=&-a^{2}+2a + 1\\=&-(2a + 1)+2a + 1\\=&0\end{aligned}$
$\begin{aligned}&2a^{2}-5a+\dfrac{1}{a}+2\\=&2(2a + 1)-5a+\dfrac{1}{\sqrt{2}+1}+2\\=&4a+2-5a+\sqrt{2}-1 + 2\\=&-a+\sqrt{2}+3\\=&-(\sqrt{2}+1)+\sqrt{2}+3\\=&2\end{aligned}$
综上，答案依次为：(1)$5$；(2)①$5$；②$0$；$2$。
