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17. (10分)运用乘法公式计算:
(1)$(1-\sqrt{5})(\sqrt{5}+1)+(\sqrt{5}-1)^{2}$;
(2)$(\sqrt{3}+\sqrt{2}-1)(\sqrt{3}-\sqrt{2}+1)$.
(1)$(1-\sqrt{5})(\sqrt{5}+1)+(\sqrt{5}-1)^{2}$;
(2)$(\sqrt{3}+\sqrt{2}-1)(\sqrt{3}-\sqrt{2}+1)$.
答案
解：
(1) $(1-\sqrt{5})(\sqrt{5}+1)+(\sqrt{5}-1)^{2}$
$=1^2 - (\sqrt{5})^2 + (\sqrt{5})^2 - 2×\sqrt{5}×1 + 1^2$
$=1 - 5 + 5 - 2\sqrt{5} + 1$
$=2 - 2\sqrt{5}$
(2) $(\sqrt{3}+\sqrt{2}-1)(\sqrt{3}-\sqrt{2}+1)$
$=[\sqrt{3}+(\sqrt{2}-1)][\sqrt{3}-(\sqrt{2}-1)]$
$=(\sqrt{3})^2 - (\sqrt{2}-1)^2$
$=3 - [(\sqrt{2})^2 - 2×\sqrt{2}×1 + 1^2]$
$=3 - (2 - 2\sqrt{2} + 1)$
$=3 - 3 + 2\sqrt{2}$
$=2\sqrt{2}$
(1) $(1-\sqrt{5})(\sqrt{5}+1)+(\sqrt{5}-1)^{2}$
$=1^2 - (\sqrt{5})^2 + (\sqrt{5})^2 - 2×\sqrt{5}×1 + 1^2$
$=1 - 5 + 5 - 2\sqrt{5} + 1$
$=2 - 2\sqrt{5}$
(2) $(\sqrt{3}+\sqrt{2}-1)(\sqrt{3}-\sqrt{2}+1)$
$=[\sqrt{3}+(\sqrt{2}-1)][\sqrt{3}-(\sqrt{2}-1)]$
$=(\sqrt{3})^2 - (\sqrt{2}-1)^2$
$=3 - [(\sqrt{2})^2 - 2×\sqrt{2}×1 + 1^2]$
$=3 - (2 - 2\sqrt{2} + 1)$
$=3 - 3 + 2\sqrt{2}$
$=2\sqrt{2}$
18. (6分)计算:$(-1)×3+\sqrt{9}+2^{2}-2025^{0}$.
答案
解：
原式 = -3 + 3 + 4 - 1
= 0 + 4 - 1
= 3
原式 = -3 + 3 + 4 - 1
= 0 + 4 - 1
= 3
