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九年级数学填空题一般
题目
小明在解决问题:已知a=12+3a=\frac{1}{2+\sqrt{3}},求2a28a+12a^{2}-8a+1的值.他是这样分析与解的:
a=12+3=23(2+3)(23)=23\because a=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3},
a2=3\therefore a-2=-\sqrt{3},
(a2)2=3\therefore \left(a-2\right)^{2}=3,a24a+4=3a^{2}-4a+4=3.
a24a=1\therefore a^{2}-4a=-1,
2a28a+1=2(a24a)+1=2×(1)+1=1\therefore 2a^{2}-8a+1=2(a^{2}-4a)+1=2\times \left(-1\right)+1=-1.
请你根据小明的分析过程,解决如下问题:
(1)(1)观察上面解答过程,请写出1n+2+n=\frac{1}{\sqrt{n+2}+\sqrt{n}}=______;
(2)(2)化简13+1+15+3+17+5++1121+119\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+⋯+\frac{1}{\sqrt{121}+\sqrt{119}}
(3)(3)a=1265a=\frac{1}{\sqrt{26}-5},请按照小明的方法求出a311a2+9a+6{a}^{3}-11{a}^{2}+9a+\sqrt{6}的值.
知识点:代数式求值、二次根式的性质与化简、二次根式的化简求值章节:未标注

答案与解析

答案

(1)1n+2+n\frac{1}{\sqrt{n+2}+\sqrt{n}}
=n+2n(n+2+n)(n+2n)=\frac{\sqrt{n+2}-\sqrt{n}}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+2}-\sqrt{n})}
=12(n+2n)=\frac{1}{2}(\sqrt{n+2}-\sqrt{n})
故答案为:12(n+2n)\frac{1}{2}(\sqrt{n+2}-\sqrt{n})
(2)13+1+15+3+17+5++1121+119(2)\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+⋯+\frac{1}{\sqrt{121}+\sqrt{119}}
=12×(31+53+75++121119)=\frac{1}{2}×(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+⋯+\sqrt{121}-\sqrt{119})
=12×(1211)=\frac{1}{2}×(\sqrt{121}-1)
=12×10=\frac{1}{2}×10
=5=5
(3)a=1265=26+5(265)(26+5)=26+5(3)\because a=\frac{1}{\sqrt{26}-5}=\frac{\sqrt{26}+5}{(\sqrt{26}-5)(\sqrt{26}+5)}=\sqrt{26}+5
a5=26\therefore a-5=\sqrt{26},即(a5)2=26\left(a-5\right)^{2}=26
a210a=1\therefore a^{2}-10a=1a3=a+10a2a^{3}=a+10a^{2}
a311a2+9a+6\therefore {a}^{3}-11{a}^{2}+9a+\sqrt{6}
=a+10a211a2+9a+6=a+10{a}^{2}-11{a}^{2}+9a+\sqrt{6}
=a2+10a+6=-{a}^{2}+10a+\sqrt{6}
=(a210a)+6=-({a}^{2}-10a)+\sqrt{6}
=1+6=-1+\sqrt{6}.

解析

(1)1n+2+n\frac{1}{\sqrt{n+2}+\sqrt{n}}
=n+2n(n+2+n)(n+2n)=\frac{\sqrt{n+2}-\sqrt{n}}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+2}-\sqrt{n})}
=12(n+2n)=\frac{1}{2}(\sqrt{n+2}-\sqrt{n})
故答案为:12(n+2n)\frac{1}{2}(\sqrt{n+2}-\sqrt{n})
(2)13+1+15+3+17+5++1121+119(2)\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+⋯+\frac{1}{\sqrt{121}+\sqrt{119}}
=12×(31+53+75++121119)=\frac{1}{2}×(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+⋯+\sqrt{121}-\sqrt{119})
=12×(1211)=\frac{1}{2}×(\sqrt{121}-1)
=12×10=\frac{1}{2}×10
=5=5
(3)a=1265=26+5(265)(26+5)=26+5(3)\because a=\frac{1}{\sqrt{26}-5}=\frac{\sqrt{26}+5}{(\sqrt{26}-5)(\sqrt{26}+5)}=\sqrt{26}+5
a5=26\therefore a-5=\sqrt{26},即(a5)2=26\left(a-5\right)^{2}=26
a210a=1\therefore a^{2}-10a=1a3=a+10a2a^{3}=a+10a^{2}
a311a2+9a+6\therefore {a}^{3}-11{a}^{2}+9a+\sqrt{6}
=a+10a211a2+9a+6=a+10{a}^{2}-11{a}^{2}+9a+\sqrt{6}
=a2+10a+6=-{a}^{2}+10a+\sqrt{6}
=(a210a)+6=-({a}^{2}-10a)+\sqrt{6}
=1+6=-1+\sqrt{6}.

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