《2025年天府前沿七年级数学上册北师大版》

7. (2025·编写)下列各式计算正确的是( )

A.$-\left|-\frac{3}{4}\right| = \frac{3}{4}$
B.$(-3) - 2 = 5$
C.$(-18) ÷ \left(-\frac{2}{3}\right) = 27$
D.$-\frac{4^{2}}{3} = \frac{16}{9}$

答案： C

8. (2025·编写)若 $|a - 3| + |2 - b| = 0$,则 $a^{2} + b^{2}$ 的值为( )

A.12
B.13
C.14
D.15

答案： B

9. (2025·编写)计算：
(1)$-\frac{1}{8} + \frac{3}{4} - \left(-\frac{1}{4}\right) - \frac{5}{8}$；
(2)$3 × (-4) + 18 ÷ (-6) - (-5)$；
(3)$\left(\frac{2}{3} - \frac{3}{4} - \frac{5}{6}\right) ÷ \left(-\frac{1}{12}\right)$；
(4)$-1.53 × 0.75 - 0.53 × \left(-\frac{3}{4}\right)$.

答案：
(1)【解】原式$=(-\frac{1}{8}-\frac{5}{8}+\frac{3}{4})+\frac{1}{4}=0+\frac{1}{4}$$=\frac{1}{4}$.
(2)【解】原式$=-12-3+5=-15+5=-10$.
(3)【解】原式$=(\frac{2}{3}-\frac{3}{4}-\frac{5}{6})× (-12)=\frac{2}{3}×$$(-12)-\frac{3}{4}× (-12)-\frac{5}{6}× (-12)=-8+9+10$$=11$.
(4)【解】原式$=-1.53× 0.75+0.53× 0.75=$$0.75× (-1.53+0.53)=0.75× (-1)=-0.75$.

10. (2025·编写)计算：
(1)$-2^{4} - (-1)^{2025} - |-9|$；
(2)$-2^{2} + 3 × (-1)^{4} - (-4) × 5$；
(3)$(-2)^{2} × \frac{1}{3} - \left(-\frac{2}{3}\right)^{3} ÷ 4$；
(4)$-8 × \left(-\frac{1}{2}\right)^{2} + \left(\frac{2}{3} - \frac{5}{8}\right) ÷ \left(-\frac{1}{24}\right)$.

答案：
(1)【解】原式$=-16-(-1)-9=-16+1-9=$$-24$.
(2)【解】原式$=-4+3× 1+20=-4+3+20=19$.
(3)【解】原式$=4× \frac{1}{3}-(-\frac{8}{27})× \frac{1}{4}=\frac{36}{27}+\frac{2}{27}=$$\frac{38}{27}$.
(4)【解】原式$=-8× \frac{1}{4}+(\frac{2}{3}-\frac{5}{8})× (-24)=$$-2+\frac{2}{3}× (-24)-\frac{5}{8}× (-24)=-2-16+15$$=-3$.