第16页

信息发布者：
D
$-16\sqrt{5}$
解：$(\sqrt{50}+\frac{1}{3}\sqrt{18}-\sqrt{162})÷\sqrt{32}$
$=(5\sqrt{2}+\frac{1}{3}×3\sqrt{2}-9\sqrt{2})÷4\sqrt{2}$
$=(5\sqrt{2}+\sqrt{2}-9\sqrt{2})÷4\sqrt{2}$
$=(-3\sqrt{2})÷4\sqrt{2}$
$=-\frac{3}{4}$
解：$(\sqrt{7}+\sqrt{5})×(\sqrt{28}-\sqrt{20})-(\sqrt{3}+3\sqrt{2})^2$
$=(\sqrt{7}+\sqrt{5})×(2\sqrt{7}-2\sqrt{5})-[(\sqrt{3})^2+2×\sqrt{3}×3\sqrt{2}+(3\sqrt{2})^2]$
$=2×(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})-(3+6\sqrt{6}+18)$
$=2×(7-5)-(21+6\sqrt{6})$
$=4-21-6\sqrt{6}$
$=-6\sqrt{6}-17$
解：$a^2-ab+b^2=(a+b)^2-3ab$
$\because a=\sqrt{5}+\sqrt{3},b=\sqrt{5}-\sqrt{3}$
$\therefore a+b=2\sqrt{5}，$$ab=(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=2$
$\therefore$原式$=(2\sqrt{5})^2-3×2=20-6=14$
解：$a^2-b^2=(a+b)(a-b)$
$\because a=\sqrt{5}+\sqrt{3},b=\sqrt{5}-\sqrt{3}$
$\therefore a+b=2\sqrt{5}，$$a-b=2\sqrt{3}$
$\therefore$原式$=2\sqrt{5}×2\sqrt{3}=4\sqrt{15}$
解：$\because a=\sqrt{2}+1$
$\therefore (a-1)^2=(\sqrt{2})^2，$即$a^2=2a+1$
将$a^2=2a+1$代入原式：
$a^3-a^2-3a+2026$
$=a(2a+1)-(2a+1)-3a+2026$
$=2a^2+a-2a-1-3a+2026$
$=2(2a+1)-4a+2025$
$=4a+2-4a+2025$
$=2027$
解：$(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\dots+\frac{1}{\sqrt{2025}+\sqrt{2026}})×(\sqrt{2026}+1)$
$=(-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-\dots-\sqrt{2025}+\sqrt{2026})×(\sqrt{2026}+1)$
$=(\sqrt{2026}-1)×(\sqrt{2026}+1)$
$=(\sqrt{2026})^2-1^2$
$=2026-1$
$=2025$
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