Задачи с корени

[fuck12]

7()/(√12) =

(5)/(√8) =

√2 ( 4 + √15 )( √10 - √6) =

√3 (√15 - √12)(√5 + 2 ) =

[ammornil]

[tex]\sqrt{a}.\sqrt{a}=a, (\forall a \ge 0)[/tex]

[tex]\sqrt{a.b}=\sqrt{a}.\sqrt{b}, \left( \left| a \ge 0 \\ b \ge 0 \right, \right)[/tex]

За рационализиране на изразите ще използваме следните преобразования:

[tex]\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{a}}.\frac{\sqrt{a}}{\sqrt{a}}=\frac{\sqrt{a}}{a} (\forall a \ge 0)[/tex]

---Примери---
1) [tex]\frac{7}{\sqrt{12}}=\frac{7}{\sqrt{4}.\sqrt{3}}=\frac{7}{2.\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}=\frac{7.\sqrt{3}}{6}[/tex]

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2) [tex]\frac{5}{\sqrt{8}}=\frac{5}{\sqrt{4}.\sqrt{2}}=\frac{5}{2.\sqrt{2}}.\frac{\sqrt{2}}{\sqrt{2}}=\frac{5.\sqrt{2}}{4}[/tex]
---
3) [tex]\sqrt{2}.(4+\sqrt{15}).(\sqrt{10}-\sqrt{6})=\sqrt{2}.(\sqrt{2.5}-\sqrt{2.3}).(4+\sqrt{15})=\sqrt{2}.(\sqrt{2}.\sqrt{5}-\sqrt{2}.\sqrt{3}).(4+\sqrt{3.5})=[/tex]

[tex]=\sqrt{2}.\sqrt{2}.(\sqrt{5}-\sqrt{3}).(4+\sqrt{3}.\sqrt{5})=2.(4.\sqrt{5}+\sqrt{5}.\sqrt{3}.\sqrt{5}-4.\sqrt{3}-\sqrt{3}.\sqrt{3}.\sqrt{5})=2.(4.\sqrt{5}+5.\sqrt{3}-4.\sqrt{3}-3.\sqrt{5})=[/tex]

[tex]=2.(\sqrt{3}+\sqrt{5})[/tex]

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4) [tex]\sqrt{3}.(\sqrt{15}-\sqrt{12}).(\sqrt{5}+2)=\sqrt{3}.(\sqrt{3.5}-\sqrt{3.4}).(\sqrt{5}+2)=\sqrt{3}.(\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{4}).(\sqrt{5}+2)=[/tex]

[tex]\sqrt{3}.\sqrt{3}.(\sqrt{5}-2).(\sqrt{5}+2)=3.\left((\sqrt{5})^{2}-2^{2}\right)=3.(5-4)=3.1=3[/tex]