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1. (营口中考) 有下列四个算式: ① $(-5)+(+3)= -8$; ② $-(-2)^3= 6$; ③ $\left(+\frac{5}{6}\right)+\left(-\frac{1}{6}\right)= \frac{2}{3}$; ④ $-3 ÷\left(-\frac{1}{3}\right)= 9$. 其中, 正确的有 ()
A.0 个
B.1 个
C.2 个
D.3 个
A.0 个
B.1 个
C.2 个
D.3 个
答案
C
2. 计算:
(1) (包头中考) $\sqrt[3]{8}+(-1)^{2024}= $____.
(2) $4^{-1}-\sqrt{\frac{1}{16}}+168^0= $____.
(3) (陕西中考) $\sqrt{25}-(-7)^0+(-2) × 3= $____.
(4) $-3^2-0.75 ÷ \frac{1}{3} ×\left[4-(-2)^3\right]=$____.
(1) (包头中考) $\sqrt[3]{8}+(-1)^{2024}= $____.
(2) $4^{-1}-\sqrt{\frac{1}{16}}+168^0= $____.
(3) (陕西中考) $\sqrt{25}-(-7)^0+(-2) × 3= $____.
(4) $-3^2-0.75 ÷ \frac{1}{3} ×\left[4-(-2)^3\right]=$____.
答案
(1) 3
(2) 1
(3) -2
(4) -36
(2) 1
(3) -2
(4) -36
3. 计算:
(1) $-1^{2025}+\left[\frac{2}{5} ×(-10)-4^2\right] ÷(-5)$.
(2) $\sqrt[3]{125}+\sqrt{\frac{25}{9}}+\sqrt[3]{-8} × \sqrt{4^2+3^2}$.
(3) $-3^2+1 ÷ 4 × \frac{1}{4}-\left|-1 \frac{1}{4}\right| ×(-0.5)^2$.
(1) $-1^{2025}+\left[\frac{2}{5} ×(-10)-4^2\right] ÷(-5)$.
(2) $\sqrt[3]{125}+\sqrt{\frac{25}{9}}+\sqrt[3]{-8} × \sqrt{4^2+3^2}$.
(3) $-3^2+1 ÷ 4 × \frac{1}{4}-\left|-1 \frac{1}{4}\right| ×(-0.5)^2$.
答案
(1) 原式 = -1 + (-4 - 16) ÷ (-5) = -1 + (-20) ÷ (-5) = -1 + 4 = 3.
(2) 原式 = 5 + $\frac{5}{3}$ - 2 × 5 = $\frac{20}{3}$ - 10 = -$\frac{10}{3}$.
(3) 原式 = -9 + $\frac{1}{16}$ - $\frac{5}{16}$ = -9$\frac{1}{4}$.
(2) 原式 = 5 + $\frac{5}{3}$ - 2 × 5 = $\frac{20}{3}$ - 10 = -$\frac{10}{3}$.
(3) 原式 = -9 + $\frac{1}{16}$ - $\frac{5}{16}$ = -9$\frac{1}{4}$.
4. 计算 $(-8) × 3 ÷(-2)^2$ 的结果为 ()
A.-6
B.6
C.-12
D.12
A.-6
B.6
C.-12
D.12
答案
A
5. 用简便方法计算 $47 ×\left(-\frac{1}{8}\right)+81 × \frac{1}{8}+26 ×(-0.125)$, 其结果是 ()
A.2
B.1
C.0
D.-1
A.2
B.1
C.0
D.-1
答案
B
