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(5+2)(52)=1(\sqrt{5}+2)(\sqrt{5}-2)=1,aa=a(a0)\sqrt{a}•\sqrt{a}=a(a≥0),(b+1)(b1)=b1(b0)(\sqrt{b}+1)(\sqrt{b}-1)=b-1(b≥0)\ldots这种两个含有二次根式的代数式相乘,积不含有二次根式,我们称这两个代数式互为有理化因式.例如5\sqrt{5}5\sqrt{5}2+1\sqrt{2}+121\sqrt{2}-123+352\sqrt{3}+3\sqrt{5}23352\sqrt{3}-3\sqrt{5}都是互为有理化因式.在进行二次根式计算时,利用有理化因式,可以化去分母中的根号,请解答下列问题:
(1)(1)化简:233\frac{2}{3\sqrt{3}}.
(2)(2)计算:123+132\frac{1}{2-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}
(3)(3)比较20242023\sqrt{2024}-\sqrt{2023}20232022\sqrt{2023}-\sqrt{2022}的大小,并说明理由.
知识点:分式的基本性质、约分、分式的加减法章节:第18章 分式 / 18.1 分式及其基本性质 / 18.1.2 分式的基本性质

答案与解析

答案

(1)233\frac{2}{3\sqrt{3}}
=2×333×3=\frac{2×\sqrt{3}}{3\sqrt{3}×\sqrt{3}}
=239=\frac{2\sqrt{3}}{9}
(2)123+132(2)\frac{1}{2-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}
=1×(2+3)(23)(2+3)+1×(3+2)(32)(3+2)=\frac{1×(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{1×(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}
=2+322(3)2+3+2(3)2(2)2=\frac{2+\sqrt{3}}{{2}^{2}-{(\sqrt{3})}^{2}}+\frac{\sqrt{3}+\sqrt{2}}{{(\sqrt{3})}^{2}-{(\sqrt{2})}^{2}}
=2+3+3+2=2+\sqrt{3}+\sqrt{3}+\sqrt{2}
=2+23+2=2+2\sqrt{3}+\sqrt{2}
(3)(3)a=20242023a=\sqrt{2024}-\sqrt{2023}b=20232022b=\sqrt{2023}-\sqrt{2022}
a>0a \gt 0b>0b \gt 0
1a=120242023=2024+2023(20242023)(2024+2023)=2024+2023\therefore \frac{1}{a}=\frac{1}{\sqrt{2024}-\sqrt{2023}}=\frac{\sqrt{2024}+\sqrt{2023}}{(\sqrt{2024}-\sqrt{2023})(\sqrt{2024}+\sqrt{2023})}=\sqrt{2024}+\sqrt{2023}1b=120232022=2023+2022(20232022)(2023+2022)=2023+2022\frac{1}{b}=\frac{1}{\sqrt{2023}-\sqrt{2022}}=\frac{\sqrt{2023}+\sqrt{2022}}{(\sqrt{2023}-\sqrt{2022})(\sqrt{2023}+\sqrt{2022})}=\sqrt{2023}+\sqrt{2022}
1a1b\therefore \frac{1}{a}>\frac{1}{b}
a>0\because a \gt 0b>0b \gt 0
a<b\therefore a \lt b
2024202320232022\therefore \sqrt{2024}-\sqrt{2023}<\sqrt{2023}-\sqrt{2022}.

解析

(1)233\frac{2}{3\sqrt{3}}
=2×333×3=\frac{2×\sqrt{3}}{3\sqrt{3}×\sqrt{3}}
=239=\frac{2\sqrt{3}}{9}
(2)123+132(2)\frac{1}{2-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}
=1×(2+3)(23)(2+3)+1×(3+2)(32)(3+2)=\frac{1×(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{1×(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}
=2+322(3)2+3+2(3)2(2)2=\frac{2+\sqrt{3}}{{2}^{2}-{(\sqrt{3})}^{2}}+\frac{\sqrt{3}+\sqrt{2}}{{(\sqrt{3})}^{2}-{(\sqrt{2})}^{2}}
=2+3+3+2=2+\sqrt{3}+\sqrt{3}+\sqrt{2}
=2+23+2=2+2\sqrt{3}+\sqrt{2}
(3)(3)a=20242023a=\sqrt{2024}-\sqrt{2023}b=20232022b=\sqrt{2023}-\sqrt{2022}
a>0a \gt 0b>0b \gt 0
1a=120242023=2024+2023(20242023)(2024+2023)=2024+2023\therefore \frac{1}{a}=\frac{1}{\sqrt{2024}-\sqrt{2023}}=\frac{\sqrt{2024}+\sqrt{2023}}{(\sqrt{2024}-\sqrt{2023})(\sqrt{2024}+\sqrt{2023})}=\sqrt{2024}+\sqrt{2023}1b=120232022=2023+2022(20232022)(2023+2022)=2023+2022\frac{1}{b}=\frac{1}{\sqrt{2023}-\sqrt{2022}}=\frac{\sqrt{2023}+\sqrt{2022}}{(\sqrt{2023}-\sqrt{2022})(\sqrt{2023}+\sqrt{2022})}=\sqrt{2023}+\sqrt{2022}
1a1b\therefore \frac{1}{a}>\frac{1}{b}
a>0\because a \gt 0b>0b \gt 0
a<b\therefore a \lt b
2024202320232022\therefore \sqrt{2024}-\sqrt{2023}<\sqrt{2023}-\sqrt{2022}.

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