第15页 - 通城学典课时作业本八年级数学人教版南通专版

$解：原式=3\sqrt {3}÷\frac {\sqrt {3}}2×2\sqrt {2}-6\sqrt {2}$

$=6×2\sqrt {2}-6\sqrt {2}$

$=6\sqrt {2}$

$解：原式=(\sqrt{7}+\sqrt{5})×(2\sqrt{7}-2\sqrt{5})-(\sqrt{3}+3\sqrt{2})^2$

$=2[(\sqrt{7})^2-(\sqrt{5})^2]-[(\sqrt{3})^2+2×\sqrt{3}×3\sqrt{2}+(3\sqrt{2})^2]$

$=2×(7-5)-[3+6\sqrt{6}+18]$

$=4-3-6\sqrt{6}-18$

$=-6\sqrt{6}-17$

$解：a=\sqrt{20}+3\sqrt{2}=2\sqrt{5}+3\sqrt{2}，b=2\sqrt{5}-\sqrt{18}=2\sqrt{5}-3\sqrt{2}$

$∴ab=(2\sqrt{5}+3\sqrt{2})(2\sqrt{5}-3\sqrt{2})=(2\sqrt{5})^2-(3\sqrt{2})^2=20-18=2$

$a+b=2\sqrt{5}+3\sqrt{2}+2\sqrt{5}-3\sqrt{2}=4\sqrt{5}$

$∴a^2-ab+b^2=(a+b)^2-3ab=(4\sqrt{5})^2-3×2=74$

$解：a=\sqrt{2}+1，a-1=\sqrt{2}$

$∴(a-1)^2=(\sqrt{2})^2，即a^2=2a+1$

$∴a^3-a^2-3a+2024=a(2a+1)-(2a+1)-3a+2024=2a^2-4a+2023=2(2a+1)-4a+2023=2025$

$解：原式=(-1+\sqrt {2}-\sqrt {2}+\sqrt {3}-\sqrt {3}+\sqrt {4}-···-\sqrt{2023}+\sqrt{2024})×(\sqrt{2024}+1)$

$=(\sqrt{2024}-1)×(\sqrt{2024}+1)$

$=2023$