1. 用简便方法计算：
(1) $(-\dfrac{1}{2})^{10} ÷ (-\dfrac{1}{2})^{-15}$；
(2) $(-1\dfrac{2}{5})^{8} × 0.25^{5} × (\dfrac{5}{7})^{8} × (-4)^{5}$；
(3) $(\dfrac{1}{10} × \dfrac{1}{9} × \dfrac{1}{8} × \dots × \dfrac{1}{2} × 1)^{10} × (10 × 9 × 8 × \dots × 2 × 1)^{10}$。

解：(1) $(-\dfrac{1}{2})^{10} ÷ (-\dfrac{1}{2})^{-15} = (-\dfrac{1}{2})^{10-(-15)} = -(\dfrac{1}{2})^{25} = -\dfrac{1}{2^{25}}.$
(2) $(-1\dfrac{2}{5})^{8} × 0.25^{5} × (\dfrac{5}{7})^{8} × (-4)^{5} = (\dfrac{7}{5})^{8} × (\dfrac{1}{4})^{5} × (\dfrac{5}{7})^{8} × (-4)^{5} = (\dfrac{7}{5} × \dfrac{5}{7})^{8} × (-4 × \dfrac{1}{4})^{5} = -1.$
(3) $(\dfrac{1}{10} × \dfrac{1}{9} × \dfrac{1}{8} × \dots × \dfrac{1}{2} × 1)^{10} × (10 × 9 × 8 × \dots × 2 × 1)^{10} = (\dfrac{1}{10} × \dfrac{1}{9} × \dfrac{1}{8} × \dots × \dfrac{1}{2} × 1 × 10 × 9 × 8 × \dots × 2 × 1)^{10} = (\dfrac{1}{10} ×10 × \dfrac{1}{9} ×9 × \dfrac{1}{8} ×8 × \dots × \dfrac{1}{2} ×2 ×1 ×1)^{10} =1^{10}=1.$

2. 用简便方法计算：
(1) $99.8 × (-100.2)$；
(2) $999^2 - 1002 × 998$；
(3) $2^3 × 5^4 + 29 × 31$；
(4) $\dfrac{2026}{2025^2 - 2026 × 2024}$。

解：(1) $99.8 × (-100.2) = - (100 - 0.2) × (100 + 0.2) = - (10\ 000 - 0.04) = -9\ 999.96.$
(2) $999^2 - 1002 × 998 = (1\ 000 - 1)^2 - (1\ 000 + 2) × (1\ 000 - 2) = 1\ 000^2 - 2\ 000 + 1 - 1\ 000^2 +4 = -1\ 995.$
(3) $2^3 × 5^4 + 29 × 31 = 2^3 × 5^3 ×5 + (30 -1) × (30 +1) = (2×5)^3 ×5 +30^2 -1 =5\ 000 +900 -1 =5\ 899.$
(4) $\dfrac{2026}{2025^2 - 2026 × 2024} = \dfrac{2\ 026}{2\ 025^2 - (2\ 025 +1) × (2\ 025 -1)} = \dfrac{2\ 026}{2\ 025^2 - (2\ 025^2 -1)} = \dfrac{2\ 026}{2\ 025^2 - 2\ 025^2 +1} =2\ 026.$

3. 教材P52习题T4变式 用简便方法计算：
(1)$57 × 99 + 44 × 99 - 99$；
(2)$55.5^2 - 44.5^2$；
(3)$-101 × 190 + 101^2 + 95^2$。

解：(1)$57 × 99 + 44 × 99 - 99 = 99 × (57 + 44 -1) =99 ×100 =9\ 900.$
(2)$55.5^2 - 44.5^2 = (55.5 +44.5) × (55.5 -44.5) =100 ×11 =1\ 100.$
(3)$-101 × 190 + 101^2 + 95^2 = 101^2 - 2 × 101 ×95 +95^2 = (101 -95)^2 =6^2 =36.$