1. (2025·烟台中考)$|-3|$的倒数是 (
A.3
B.$\frac{1}{3}$
C.$-3$
D.$-\frac{1}{3}$
B
)A.3
B.$\frac{1}{3}$
C.$-3$
D.$-\frac{1}{3}$
答案
1. B 解析:因为$|-3|=3,3$的倒数是$\frac{1}{3}$,所以$|-3|$的倒数是$\frac{1}{3}$,故选 B.
2. $99\frac{18}{19}×13=(100-\frac{1}{19})×13=1300-\frac{13}{19}$运用了(
A.加法交换律
B.乘法结合律
C.乘法分配律
D.乘法交换律和结合律
C
)A.加法交换律
B.乘法结合律
C.乘法分配律
D.乘法交换律和结合律
答案
2. C 解析:括号中的两项分别乘括号外的项,运用了乘法分配律.故选 C.
3.(济南中考)下列计算$(-55)×99+(-44)×99-99$正确的是(
A.原式$=99×(-55-44)=-9801$
B.原式$=99×(-55-44+1)=-9702$
C.原式$=99×(-55-44-1)=-9900$
D.原式$=99×(-55-44-99)=-19602$
C
)A.原式$=99×(-55-44)=-9801$
B.原式$=99×(-55-44+1)=-9702$
C.原式$=99×(-55-44-1)=-9900$
D.原式$=99×(-55-44-99)=-19602$
答案
3. C 解析:利用乘法分配律将 99 提出,则原式$=99×(-55-44-1)=-9900$.故选 C.
4. 绝对值与倒数均等于它本身的数是
1
.答案
4. 1 解析:绝对值等于它本身的数是非负数,倒数等于它本身的数有 1 和-1,故满足题意的数是 1.
5. 在式子“$3× □ -2× □ =15$”中的两个方框内分别填入一个数,使这两个数互为相反数且式子成立,则第一个方框内的数是
3
.答案
5. 3 解析:设第一个
6. 用乘法运算律先将下列算式变形,再计算.
(1) $8×(-5.06)×1.25 =$
$=$
(2) $[(-\dfrac{5}{6})×4]×(-1\dfrac{1}{5}) =$
$=$
(3) $\dfrac{4}{5}×(-\dfrac{5}{13}) - (-\dfrac{3}{5})×(-\dfrac{5}{13}) - \dfrac{5}{13}×(-1\dfrac{3}{5}) =$
$=$
(1) $8×(-5.06)×1.25 =$
$8×1.25×(-5.06)$
$=$
$-50.6$
;(2) $[(-\dfrac{5}{6})×4]×(-1\dfrac{1}{5}) =$
$(-\dfrac{5}{6})×(-\dfrac{6}{5})×4$
$=$
$4$
;(3) $\dfrac{4}{5}×(-\dfrac{5}{13}) - (-\dfrac{3}{5})×(-\dfrac{5}{13}) - \dfrac{5}{13}×(-1\dfrac{3}{5}) =$
$(-\dfrac{5}{13})×[\dfrac{4}{5}-(-\dfrac{3}{5})+(-\dfrac{8}{5})]$
$=$
$\dfrac{1}{13}$
.答案
6. (1)$8×1.25×(-5.06)$ $-50.6$
(2) $(-\dfrac{5}{6})×(-\dfrac{6}{5})×4$ $4$
(3) $(-\dfrac{5}{13})×[\dfrac{4}{5}-(-\dfrac{3}{5})+(-\dfrac{8}{5})]$ $\dfrac{1}{13}$
(2) $(-\dfrac{5}{6})×(-\dfrac{6}{5})×4$ $4$
(3) $(-\dfrac{5}{13})×[\dfrac{4}{5}-(-\dfrac{3}{5})+(-\dfrac{8}{5})]$ $\dfrac{1}{13}$
7. 教材P47练习T1变式 计算:
(1) $-2×(-8)×5×(-1\dfrac{1}{4})$;
(2) $(-0.25)×0.5×(-\dfrac{2}{7})×4$;
(3) $(-56)×(\dfrac{4}{7}-\dfrac{3}{8}+\dfrac{1}{14})$;
(4) $(-3.59)×(-\dfrac{4}{7})-2.41×(-\dfrac{4}{7})+6×(-\dfrac{4}{7})$。
(1) $-2×(-8)×5×(-1\dfrac{1}{4})$;
(2) $(-0.25)×0.5×(-\dfrac{2}{7})×4$;
(3) $(-56)×(\dfrac{4}{7}-\dfrac{3}{8}+\dfrac{1}{14})$;
(4) $(-3.59)×(-\dfrac{4}{7})-2.41×(-\dfrac{4}{7})+6×(-\dfrac{4}{7})$。
答案
7. (1) $-2×(-8)×5×(-1\dfrac{1}{4}) = (-2×5)×[(-8)×(-\dfrac{5}{4})] = -10×10 = -100.$
(2) $(-0.25)×0.5×(-\dfrac{2}{7})×4 = (-0.25)×4×\dfrac{1}{2}×(-\dfrac{2}{7}) = -\dfrac{1}{2}×(-\dfrac{2}{7}) = \dfrac{1}{7}.$
(3) $(-56)×(\dfrac{4}{7}-\dfrac{3}{8}+\dfrac{1}{14}) = (-56)×\dfrac{4}{7}-(-56)×\dfrac{3}{8}+(-56)×\dfrac{1}{14} = -32+21-4 = -15.$
(4) $(-3.59)×(-\dfrac{4}{7})-2.41×(-\dfrac{4}{7})+6×(-\dfrac{4}{7}) = (-3.59-2.41+6)×(-\dfrac{4}{7}) = 0×(-\dfrac{4}{7}) = 0.$
(2) $(-0.25)×0.5×(-\dfrac{2}{7})×4 = (-0.25)×4×\dfrac{1}{2}×(-\dfrac{2}{7}) = -\dfrac{1}{2}×(-\dfrac{2}{7}) = \dfrac{1}{7}.$
(3) $(-56)×(\dfrac{4}{7}-\dfrac{3}{8}+\dfrac{1}{14}) = (-56)×\dfrac{4}{7}-(-56)×\dfrac{3}{8}+(-56)×\dfrac{1}{14} = -32+21-4 = -15.$
(4) $(-3.59)×(-\dfrac{4}{7})-2.41×(-\dfrac{4}{7})+6×(-\dfrac{4}{7}) = (-3.59-2.41+6)×(-\dfrac{4}{7}) = 0×(-\dfrac{4}{7}) = 0.$
8. 下列说法:①互为倒数的两个数相乘,积为1;②正数的倒数是正数,负数的倒数是负数;③小于-1的数的倒数大于其本身;④大于1的数的倒数小于其本身.其中正确的有(
A.1个
B.2个
C.3个
D.4个
D
)A.1个
B.2个
C.3个
D.4个
答案
8. D 解析:①②③④说法均正确,有4个.故选 D.
9. 算式$743×369 -741×370$的值是(
A.-3
B.-2
C.2
D.3
A
)A.-3
B.-2
C.2
D.3
答案
9. A 解析:$743×369-741×370 = 743×369-741×(369+1) = 743×369-741×369-741 = (743-741)×369-741 = 738-741 = -3.$故选 A.
10. 若两个数的积为-1,我们称它们互为负倒数,则0.25的负倒数是
-4
。答案
10. -4 解析:因为$-1÷0.25=-4$,所以 0.25 的负倒数是-4.
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