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初中数学七年级上册(581题)


第121题

请你判断小明的解答是否正确,并说明理由.



参考答案:正确,
理由为:一个数的倒数的倒数等于原数;


第122题

请你运用小明的解法解答下面的问题.




计算:\(( - \frac{1}{{24}}) \div (\frac{1}{3} - \frac{1}{6} + \frac{3}{8})\)



参考答案:原式的倒数为\((\frac{1}{3} - \frac{1}{6} + \frac{3}{8}) \div ( - \frac{1}{{24}}) = (\frac{1}{3} - \frac{1}{6} + \frac{3}{8}) \times ( - 24) = - 8 + 4 - 9 = - 13\),则\(( - \frac{1}{{24}}) \div (\frac{1}{3} - \frac{1}{6} + \frac{3}{8}) = - \frac{1}{{13}}\).



第124题

\({a^2} = 25\)\(|b| = 3\),则\(a + b\)所有可能的值为\((\)  \()\).


A.8

B.8或2

C.8或\( - 2\)

D.\( \pm 8\)或\( \pm 2\)


参考答案:D



第126题

下列算式中,运算结果为负数的是\((\)  \()\).


A.\( - ( - 5)\)

B.\(| - 5|\)

C.\({( - 5)^3}\)

D.\({( - 5)^2}\)


参考答案:C


第127题

\({( - \frac{1}{2})^2} = \)___



参考答案:\(\frac{1}{4}\).


第128题

已知\(a\)\(b\)互为相反数,\(c\)\(d\)互为倒数,\(x\)的绝对值是2,试求:\({x^2} - {(a + b)^{2021}} + {( - cd)^{2022}}\)



参考答案:解:由已知可得\(a + b = 0\),\(cd = 1\),\(x = \pm 2\).

当\(x = 2\)时,原式\( = {2^2} - (0 + 1) \times 2 + {0^{2007}} + {( - 1)^{2008}} = 4 - 2 + 1 = 3\);


当\(x = - 2\)时,原式\( = {( - 2)^2} - (0 + 1) \times ( - 2) + {0^{2007}} + {( - 1)^{2008}} = 4 + 2 + 1 = 7\).


第129题

已知\(|x + 1| = 4\)\({(y + 2)^2} = 4\),求\(x + y\)的值.



参考答案:解:\(\because |x + 1| = 4\),\({(y + 2)^2} = 4\),\(\therefore x + 1 = 4\),或\(x + 1 = - 4\),\(y + 2 = 2\)或\(y + 2 = - 2\),解得\(x = 3\)或\(x = - 5\),\(y = 0\)或\(y = - 4\),\(\therefore x = 3\),\(y = 0\)时,\(x + y = 3 + 0 = 3\);\(x = 3\),\(y = - 4\)时,\(x + y = 3 - 4 = - 1\);\(x = - 5\),\(y = 0\)时,\(x + y = - 5 + 0 = - 5\);\(x = - 5\),\(y = - 4\)时,\(x + y = - 5 - 4 = - 9\).


第130题

计算:\(76 \times 20.22 + 43 \times 20.22 - 19 \times 20.22 = \)___



参考答案:2022


解析:

解:原式\( = 20.22 \times (76 + 43 - 19)\)

\( = 20.22 \times 100\)

\( = 2022\)

故答案为:2022.


第131题

计算:\([ - 3 \times {( - \frac{1}{3})^2} + {( - 1)^3}] \div ( - \frac{2}{3})\)



参考答案:解:原式\( = [ - 3 \times \frac{1}{9} + ( - 1)] \times ( - \frac{3}{2})\)
\( = [ - \frac{1}{3} + ( - 1)] \times ( - \frac{3}{2})\)
\( = - \frac{4}{3} \times ( - \frac{3}{2})\)
\( = 2\).




第134题

符号“\(f\)”表示一种运算,它对一些数的运算如下:\(f(1) = 1 - \frac{1}{2}\)\(f(2) = 1 - \frac{1}{3}\)\(f(3) = 1 - \frac{1}{4}\)\(f(4) = 1 - \frac{1}{5}\)\( \ldots \)利用以上运算的规律,写出\(f(n) = \)___\((n\)为正整数),计算\(f\left ( {1} \right )\)\(\cdot f\left ( {2} \right )\)\(\cdot f\left ( {3} \right )\)\( \cdot \ldots \cdot f(100) = \)___



参考答案:\(1 - \frac{1}{{n + 1}}\);
\(\frac{1}{{101}}\).


解析:

解:(1\(\because \)\(f(1) = 1 - \frac{1}{2}\)\(f(2) = 1 - \frac{1}{3}\)\(f(3) = 1 - \frac{1}{4}\)\(f(4) = 1 - \frac{1}{5}\)\( \ldots \)

\(\therefore f(n) = 1 - \frac{1}{{n + 1}}\)

\(f\)1\( \cdot f\)2\( \cdot f\)3\( \cdot \ldots \cdot f(100)\)

\( = (1 - \frac{1}{2})(1 - \frac{1}{3})(1 - \frac{1}{4}) \cdot \cdot \cdot (1 - \frac{1}{{101}})\)

\( = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdot \cdot \cdot \times \frac{{100}}{{101}}\)

\( = \frac{1}{{101}}\)

故答案为:\(1 - \frac{1}{{n + 1}}\)\(\frac{1}{{101}}\)


第135题

用四舍五入法按要求对0.05019分别取近似值,其中错误的是\((\)  \()\).


A.0.1(精确到\(0.1)\)

B.0.05(精确到百分位)

C.0.05(精确到千分位)

D.0.0502(精确到\(0.0001)\)


参考答案:C




第138题

数对\(( - 2,1)\)\((3, - \frac{1}{2})\)中是“共生有理数对”的是___.



参考答案:无;


解析:

\( - 2 - 1 = - 3\)\( - 2 \times 1 + 1 = - 1\)

\(\therefore - 2 - 1 \ne - 2 \times 1 + 1\)

\(\therefore ( - 2,1)\)不是“共生有理数对”

\(\because 3 - ( - \frac{1}{2}) = \frac{7}{2}\)\(3 \times ( - \frac{1}{2}) + 1 = - \frac{1}{2}\)

\(\therefore 3 - ( - \frac{1}{2}) \ne 3 \times ( - \frac{1}{2}) + 1\)

\(\therefore (3, - \frac{1}{2})\)不是“共生有理数对”


第139题

\((a,3)\)是“共生有理数对”,求\(a\)的值.



参考答案:由题意得:\(a - 3 = 3a + 1\)解得\(a = - 2\).


解析:

\( - 2 - 1 = - 3\)\( - 2 \times 1 + 1 = - 1\)

\(\therefore - 2 - 1 \ne - 2 \times 1 + 1\)

\(\therefore ( - 2,1)\)不是“共生有理数对”

\(\because 3 - ( - \frac{1}{2}) = \frac{7}{2}\)\(3 \times ( - \frac{1}{2}) + 1 = - \frac{1}{2}\)

\(\therefore 3 - ( - \frac{1}{2}) \ne 3 \times ( - \frac{1}{2}) + 1\)

\(\therefore (3, - \frac{1}{2})\)不是“共生有理数对”


第140题

白兔的只数200只,灰兔有多少只?符合列式\(200 \times (1 + \frac{1}{{10}})\)的条件是\((\)  \()\)


A.灰兔的只数是白兔的只数\(\frac{1}{{10}}\)

B.白兔的只数是灰兔的只数\(\frac{1}{{10}}\)

C.白兔的只数比灰兔的只数多\(\frac{1}{{10}}\)

D.灰兔的只数比白兔的只数多\(\frac{1}{{10}}\)


参考答案:D


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