《2025年名校课堂七年级数学上册北师大版》

计算：
(1)$-0.75×(-\frac {3}{2})÷(-\frac {9}{4}).$
(2)$3×(-1)+2^{2}+|-4|.$
(3)$(-1)×(-4)+2^{2}÷(7-5).$
(4)$-2^{2}×(-\frac {3}{4})-(-3)^{3}÷9.$
(7)$-2^{2}×7-(-3)×6-5÷(-\frac {1}{5}).$
(8)$[(-1)^{100}+(1-\frac {1}{2})×\frac {1}{3}]÷(-3^{2}+2).$
(9)$(\frac {7}{9}-\frac {5}{6}+\frac {7}{18})×2×3^{2}-\frac {7}{4}÷(-1.75).$
(10)$-1^{2026}-(1-0×4)÷\frac {1}{3}×[(-2)^{2}-6].$

答案：
(1)原式= $-\frac{3}{4} × (-\frac{3}{2}) × (-\frac{4}{9}) = -\frac{1}{2}$.
(2)原式= $-3 + 4 + 4 = 5$.
(3)原式= $(-1) × (-4) + 4 ÷ 2 = 4 + 2 = 6$.
(4)原式= $-4 × (-\frac{3}{4}) + 27 ÷ 9 = 3 + 3 = 6$.
(7)原式= $-4 × 7 + 18 + 5 × 5 = -28 + 18 + 25 = 15$.
(8)原式= $(1 + \frac{1}{2} × \frac{1}{3}) ÷ (-9 + 2) = (1 + \frac{1}{6}) ÷ (-7) = \frac{7}{6} × (-\frac{1}{7}) = -\frac{1}{6}$.
(9)原式= $(\frac{7}{9} - \frac{5}{6} + \frac{7}{18}) × (2 × 9) - \frac{7}{4} × (-\frac{4}{7}) = (\frac{7}{9} - \frac{5}{6} + \frac{7}{18}) × 18 + 1 = \frac{7}{9} × 18 - \frac{5}{6} × 18 + \frac{7}{18} × 18 + 1 = 14 - 15 + 7 + 1 = 7$.
(10)原式= $-1 - (1 - 0) × 3 × (4 - 6) = -1 - 1 × 3 × (-2) = -1 + 6 = 5$.