9. 下列计算正确的是 (
A.$\sqrt{6}÷(\sqrt{3}-\sqrt{2})=\sqrt{2}-\sqrt{3}$
B.$\sqrt{(-9)×(-25)}=\sqrt{-9}×\sqrt{-25}=(-3)×(-5)=15$
C.$\sqrt{2}×(\sqrt{3}+\sqrt{2})=\sqrt{10}$
D.$\sqrt{13^{2}-12^{2}}=\sqrt{(13+12)×(13-12)}=5$
D
)A.$\sqrt{6}÷(\sqrt{3}-\sqrt{2})=\sqrt{2}-\sqrt{3}$
B.$\sqrt{(-9)×(-25)}=\sqrt{-9}×\sqrt{-25}=(-3)×(-5)=15$
C.$\sqrt{2}×(\sqrt{3}+\sqrt{2})=\sqrt{10}$
D.$\sqrt{13^{2}-12^{2}}=\sqrt{(13+12)×(13-12)}=5$
答案：9.D
10. (2024·启东期末)已知$a=4+2\sqrt{5}$，$b=4-2\sqrt{5}$，则$a^{2}b-ab^{2}$的值为
$-16\sqrt{5}$
.答案：10.$-16\sqrt{5}$
解析：
解：$a^{2}b - ab^{2} = ab(a - b)$
$a = 4 + 2\sqrt{5}$，$b = 4 - 2\sqrt{5}$
$ab=(4 + 2\sqrt{5})(4 - 2\sqrt{5})=16 - (2\sqrt{5})^{2}=16 - 20=-4$
$a - b=(4 + 2\sqrt{5}) - (4 - 2\sqrt{5})=4\sqrt{5}$
$ab(a - b)=-4×4\sqrt{5}=-16\sqrt{5}$
$-16\sqrt{5}$
$a = 4 + 2\sqrt{5}$，$b = 4 - 2\sqrt{5}$
$ab=(4 + 2\sqrt{5})(4 - 2\sqrt{5})=16 - (2\sqrt{5})^{2}=16 - 20=-4$
$a - b=(4 + 2\sqrt{5}) - (4 - 2\sqrt{5})=4\sqrt{5}$
$ab(a - b)=-4×4\sqrt{5}=-16\sqrt{5}$
$-16\sqrt{5}$
11. 计算:
(1) $(\sqrt{50}+\frac{1}{3}\sqrt{18}-\sqrt{162})÷\sqrt{32}$；
(2) $(\sqrt{7}+\sqrt{5})×(\sqrt{28}-\sqrt{20})-(\sqrt{3}+3\sqrt{2})^{2}$.
(1) $(\sqrt{50}+\frac{1}{3}\sqrt{18}-\sqrt{162})÷\sqrt{32}$；
(2) $(\sqrt{7}+\sqrt{5})×(\sqrt{28}-\sqrt{20})-(\sqrt{3}+3\sqrt{2})^{2}$.
答案：11.(1)$-\frac{3}{4}$ (2)$-6\sqrt{6}-17$
解析：
(1)原式$=(5\sqrt{2}+\frac{1}{3}×3\sqrt{2}-9\sqrt{2})÷4\sqrt{2}$
$=(5\sqrt{2}+\sqrt{2}-9\sqrt{2})÷4\sqrt{2}$
$=(-3\sqrt{2})÷4\sqrt{2}$
$=-\frac{3}{4}$
(2)原式$=(\sqrt{7}+\sqrt{5})×(2\sqrt{7}-2\sqrt{5})-(\sqrt{3}+3\sqrt{2})^{2}$
$=2(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})-(3 + 6\sqrt{6}+18)$
$=2×(7 - 5)-(21 + 6\sqrt{6})$
$=2×2 - 21 - 6\sqrt{6}$
$=4 - 21 - 6\sqrt{6}$
$=-17 - 6\sqrt{6}$
$=(5\sqrt{2}+\sqrt{2}-9\sqrt{2})÷4\sqrt{2}$
$=(-3\sqrt{2})÷4\sqrt{2}$
$=-\frac{3}{4}$
(2)原式$=(\sqrt{7}+\sqrt{5})×(2\sqrt{7}-2\sqrt{5})-(\sqrt{3}+3\sqrt{2})^{2}$
$=2(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})-(3 + 6\sqrt{6}+18)$
$=2×(7 - 5)-(21 + 6\sqrt{6})$
$=2×2 - 21 - 6\sqrt{6}$
$=4 - 21 - 6\sqrt{6}$
$=-17 - 6\sqrt{6}$
12. (教材变式)已知$a=\sqrt{5}+\sqrt{3}$，$b=\sqrt{5}-\sqrt{3}$，求下面的代数式的值:
(1) $a^{2}-ab+b^{2}$；
(2) $a^{2}-b^{2}$.
(1) $a^{2}-ab+b^{2}$；
(2) $a^{2}-b^{2}$.
答案：12.(1)原式=$(a + b)^2 - 3ab.\because a = \sqrt{5} + \sqrt{3},b = \sqrt{5} - \sqrt{3},\therefore a + b = 2\sqrt{5},ab = 2.\therefore$原式=$(2\sqrt{5})^2 - 3×2 = 14$ (2)由题意及(1)，得$a + b = 2\sqrt{5},a - b = 2\sqrt{3}.\therefore$原式=$(a + b)(a - b)=2\sqrt{5}×2\sqrt{3}=4\sqrt{15}$
解析：
(1) $a^{2}-ab+b^{2}=(a + b)^2 - 3ab$，
$\because a = \sqrt{5} + \sqrt{3},b = \sqrt{5} - \sqrt{3}$，
$\therefore a + b = (\sqrt{5} + \sqrt{3}) + (\sqrt{5} - \sqrt{3}) = 2\sqrt{5}$，
$ab = (\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2$，
$\therefore$原式$=(2\sqrt{5})^2 - 3×2 = 20 - 6 = 14$；
(2) $a^{2}-b^{2}=(a + b)(a - b)$，
$\because a + b = 2\sqrt{5}$，$a - b = (\sqrt{5} + \sqrt{3}) - (\sqrt{5} - \sqrt{3}) = 2\sqrt{3}$，
$\therefore$原式$=2\sqrt{5}×2\sqrt{3}=4\sqrt{15}$.
$\because a = \sqrt{5} + \sqrt{3},b = \sqrt{5} - \sqrt{3}$，
$\therefore a + b = (\sqrt{5} + \sqrt{3}) + (\sqrt{5} - \sqrt{3}) = 2\sqrt{5}$，
$ab = (\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2$，
$\therefore$原式$=(2\sqrt{5})^2 - 3×2 = 20 - 6 = 14$；
(2) $a^{2}-b^{2}=(a + b)(a - b)$，
$\because a + b = 2\sqrt{5}$，$a - b = (\sqrt{5} + \sqrt{3}) - (\sqrt{5} - \sqrt{3}) = 2\sqrt{3}$，
$\therefore$原式$=2\sqrt{5}×2\sqrt{3}=4\sqrt{15}$.
13. 已知$a=\sqrt{2}+1$，求$a^{3}-a^{2}-3a+2026$的值.
答案：13.$\because a = \sqrt{2}+1,\therefore(a - 1)^2 = (\sqrt{2})^2$，即$a^2 = 2a + 1.\therefore$原式=$a(2a + 1)-(2a + 1)-3a + 2026 = 2a^2-4a + 2025 = 2(2a + 1)-4a + 2025 = 2027$
解析：
$\because a = \sqrt{2} + 1$，
$\therefore (a - 1)^2 = (\sqrt{2})^2$，
即$a^2 - 2a + 1 = 2$，
$\therefore a^2 = 2a + 1$。
原式$=a^3 - a^2 - 3a + 2026$
$=a · a^2 - a^2 - 3a + 2026$
$=a(2a + 1) - (2a + 1) - 3a + 2026$
$=2a^2 + a - 2a - 1 - 3a + 2026$
$=2a^2 - 4a + 2025$
$=2(2a + 1) - 4a + 2025$
$=4a + 2 - 4a + 2025$
$=2027$
$\therefore (a - 1)^2 = (\sqrt{2})^2$，
即$a^2 - 2a + 1 = 2$，
$\therefore a^2 = 2a + 1$。
原式$=a^3 - a^2 - 3a + 2026$
$=a · a^2 - a^2 - 3a + 2026$
$=a(2a + 1) - (2a + 1) - 3a + 2026$
$=2a^2 + a - 2a - 1 - 3a + 2026$
$=2a^2 - 4a + 2025$
$=2(2a + 1) - 4a + 2025$
$=4a + 2 - 4a + 2025$
$=2027$
14. 观察下列各式的计算过程，并寻找规律:
$\frac{1}{1+\sqrt{2}}=\frac{1-\sqrt{2}}{(1+\sqrt{2})×(1-\sqrt{2})}=-1+\sqrt{2}$；
$\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}+\sqrt{3})×(\sqrt{2}-\sqrt{3})}=-\sqrt{2}+\sqrt{3}$；
$\frac{1}{\sqrt{3}+\sqrt{4}}=\frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3}+\sqrt{4})×(\sqrt{3}-\sqrt{4})}=-\sqrt{3}+\sqrt{4}$；
…
利用发现的规律计算:$(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+···+\frac{1}{\sqrt{2025}+\sqrt{2026}})×(\sqrt{2026}+1)$.
$\frac{1}{1+\sqrt{2}}=\frac{1-\sqrt{2}}{(1+\sqrt{2})×(1-\sqrt{2})}=-1+\sqrt{2}$；
$\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}+\sqrt{3})×(\sqrt{2}-\sqrt{3})}=-\sqrt{2}+\sqrt{3}$；
$\frac{1}{\sqrt{3}+\sqrt{4}}=\frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3}+\sqrt{4})×(\sqrt{3}-\sqrt{4})}=-\sqrt{3}+\sqrt{4}$；
…
利用发现的规律计算:$(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+···+\frac{1}{\sqrt{2025}+\sqrt{2026}})×(\sqrt{2026}+1)$.
答案：14.原式=$(-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-···-\sqrt{2025}+\sqrt{2026})×(\sqrt{2026}+1)=(\sqrt{2026}-1)×(\sqrt{2026}+1)=2025$