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初中数学八年级上册试题库(555题)


第241题

如图,MON=60°,点A在射线OM上,点BC在射线ON上(点C在点B的右侧),且OAB+OAC=60°.点B关于直线OM的对称点为D,连接CD.用等式表示线段CDAB之间的数量关系,并证明。

图片 11



参考答案:\(CD = AB\)证明:连接\(AD\),\(OD\),\(\because \)点\(B\)关于直线\(OM\)的对称点为\(D\),点\(A\)在射线\(OM\)上,\(\therefore AD = AB\),\(\angle OAD = \angle OAB\),\(\because \angle OAB + \angle OAC = 60^\circ \),\(\therefore \angle OAD + \angle OAC = 60^\circ \),即\(\angle DAC = 60^\circ \),在\(\vartriangle OAC\)中,\(\angle ACO = 180^\circ - \angle OAC - \angle AOC = 180^\circ - (60^\circ - \angle OAB) - 60^\circ = 60^\circ + \angle OAB\),又\(\because \angle ABC = \angle AOB + \angle OAB = 60^\circ + \angle OAB\),\(\therefore \angle ACO = \angle ABC\),\(\therefore AB = AC\),\(\therefore AC = AD\),又\(\because \angle CAD = 60^\circ \),\(\therefore \vartriangle ACD\)是等边三角形,\(\therefore CD = AD\),\(\therefore CD = AB\).


第242题

AEC的度数;



参考答案:60°


第243题

\(0^\circ < \alpha < 60^\circ \)时,用等式表示线段AECDDE之间的数量关系,并证明。



参考答案:


\(CD = 2DE + AE\)



证明:如图,在\(CD\)上截取\(BG = BE\)





\(\because \angle BEC = 60^\circ \)



\(\therefore \Delta BGE\)是等边三角形,



\(\therefore \angle BGC = \angle AED = 120^\circ \)



\(\because \angle BCE = \angle DAE = \alpha \)



\(\therefore \Delta BCG \cong \Delta DAE(AAS)\)



\(\therefore AE = CG\)



\(\because EG = BE = DE\)



\(\therefore CD = 2DE + CG\)



\(CD = 2DE + AE\)









第250题

\( (ab{)}^{5}\)



参考答案:\( {a}^{5}{b}^{5}\)


第251题

\({( - 4{x^3})^2}\)



参考答案:\( 16{x}^{6}\)


第252题

\( {\left(-3x\right)}^{3}\)



参考答案:\( -27{x}^{3}\)


第253题

计算:\({0.75^{2022}} \times {( - \frac{4}{3})^{2021}}\)的结果为\((\)  \()\)


A.\(\frac{4}{3}\)

B.\(\frac{3}{4}\)

C.\( - \frac{4}{3}\)

D.\( - \frac{3}{4}\)


参考答案:D


第254题

\( {\left(-4\times 1{0}^{3}\right)}^{2}\)            



参考答案:\( 1.6\times 1{0}^{7}\)


第255题

\( (5a+4b)\left(5a-4b-3\right)\) 



参考答案:8


第256题

下列运算正确的是\((\)  \()\)


A.\({(ab)^2} = {a^2}{b^2}\)

B.\({a^3} + {a^2} = {a^5}\)

C.\({a^3} \cdot {a^2} = {a^6}\)

D.\( (a+b{)}^{2}={a}^{2}+{b}^{2}\)


参考答案:A






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