Nielamaeidn
Answered

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please answer my question
1. √2(√3-√5)
2. 4√5(√5+3√15)
3. √3(√21-√12+√24
4. (2√5+5√3)²
5. (√3+√8)(√3-√8)
and
1. [tex] \frac{3}{ \sqrt{5} } [/tex]
2.[tex] \frac{8}{ \sqrt[3]{16} } [/tex]
3. [tex] \frac{4}{1- \sqrt{2} } [/tex]
4. [tex] \frac{ \sqrt{3} + \sqrt{5} }{ \sqrt{5} - \sqrt{3} } [/tex]
5. [tex] \frac{ \sqrt{3}- 4 }{ \sqrt{2}+ \sqrt{3} } [/tex]

Sagot :

Riza1idn
[tex]1)\\\\ \sqrt{2}( \sqrt{3}- \sqrt{5})= \sqrt{2}\cdot \sqrt{3}- \sqrt{2}\cdot \sqrt{5}=\sqrt{2\cdot 3}-\sqrt{2\cdot 5}=\sqrt{6}-\sqrt{10}\\\\2)\\\\ 4\sqrt{ 5}( \sqrt{5}+3 \sqrt{15})= 4\sqrt{ 5}\cdot \sqrt{5}+4\sqrt{ 5}\cdot 3 \sqrt{15}=4\cdot\sqrt{ 5\cdot 5}+12\cdot \sqrt{5\cdot 15}=\\\\=4\cdot\sqrt{ 25}+12\cdot \sqrt{75}=4\cdot 5+12\cdot \sqrt{25*3}=20+12\cdot \sqrt{25 }\cdot \sqrt{3}=20+12\cdot 5\cdot \sqrt{ 3}=\\\\=20+60\sqrt{ 3}[/tex]

[tex]3)\\\\ \sqrt{3}(\sqrt{ 21}- \sqrt{12}+ \sqrt{24 })= \sqrt{3} \cdot \sqrt{ 21}-\sqrt{3} \cdot \sqrt{12}+\sqrt{3} \cdot \sqrt{24 }=\sqrt{ 21\cdot 3}- \sqrt{12\cdot 3}+ \sqrt{24\cdot 3 }=\\\\=\sqrt{ 63 }- \sqrt{36}+ \sqrt{72}=\sqrt{9\cdot 7 }-6+ \sqrt{36*2}=\sqrt{9 }\cdot \sqrt{7}+ \sqrt{36 }\sqrt{2}-6=\\\\=3\sqrt{7}+6\sqrt{2}-6[/tex]

[tex]4)\\\\(2\sqrt{5}+5 \sqrt{3})^2=(2\sqrt{5})^2+2\cdot 2\sqrt{5} \cdot 5 \sqrt{3}+(5 \sqrt{3})^2=4*5+20\sqrt{5\cdot 3}+25*3=\\\\=20+20\sqrt{15}+75= 95+20\sqrt{15}[/tex]

[tex]5)\\\\ ( \sqrt{3}+ \sqrt{8})( \sqrt{3}- \sqrt{8}) =(\sqrt{3})^2- (\sqrt{8})^2=3-8=-5[/tex]


[tex]1)\\\\ \frac{3}{ \sqrt{5} } = \frac{3}{ \sqrt{5} }\cdot \frac{ \sqrt{5}}{ \sqrt{5} }=\frac{3\sqrt{5}}{\sqrt{5\cdot 5}}=\frac{3\sqrt{5}}{\sqrt{25}}= \frac{3\sqrt{5}}{ 5 }\\\\2)\\\\ \frac{8}{ \sqrt[3]{16} } = \frac{8}{ \sqrt[3]{16} } \cdot \frac{ \sqrt[3]{4}}{ \sqrt[3]{4} } =\frac{8\sqrt[3]{4}}{ \sqrt[3]{16\cdot 4} }=\frac{8\sqrt[3]{4}}{ \sqrt[3]{64}}=\frac{8\sqrt[3]{4}}{ 4} =2\sqrt[3]{4}[/tex]

[tex]3)\\\\ \frac{4}{1- \sqrt{2} } =\frac{4}{1- \sqrt{2} } \cdot \frac{1+\sqrt{2}}{1+ \sqrt{2} }= \frac{4(1+\sqrt{2})}{1^2- (\sqrt{2})^2 }=\frac{4(1+\sqrt{2})}{1-2 }=\frac{4(1+\sqrt{2})}{-1 }=-4(1+\sqrt{2})[/tex]

[tex]4)\\\\ \frac{ \sqrt{3} + \sqrt{5} }{ \sqrt{5} - \sqrt{3} } = \frac{ \sqrt{3} + \sqrt{5} }{ \sqrt{5} - \sqrt{3} } \cdot \frac{\sqrt{5}+ \sqrt{3} }{ \sqrt{5}+ \sqrt{3} }=\frac{(\sqrt{3}+ \sqrt{5})^2 }{(\sqrt{5})^2-(\sqrt{3})^2}= \frac{(\sqrt{3})^2+2\cdot \sqrt{3} \cdot \sqrt{5}+ (\sqrt{5})^2 }{5-3}= \\\\= \frac{3+2\cdot \sqrt{3\cdot 5} + 5 }{2}= \frac{8+2 \sqrt{15} }{2}= \frac{2(4+ \sqrt{15} )}{2}=4+ \sqrt{15}[/tex]

[tex]5)\\\\ \frac{ \sqrt{3}- 4 }{ \sqrt{2}+ \sqrt{3} } = \frac{ \sqrt{3}- 4 }{ \sqrt{2}+ \sqrt{3} }\cdot \frac{ \sqrt{2}- \sqrt{3} }{ \sqrt{2}- \sqrt{3} } =\frac{\sqrt{3}\cdot \sqrt{2}-\sqrt{3}\cdot \sqrt{3}-4\sqrt{2}+4\sqrt{3}}{( \sqrt{2})^2- (\sqrt{3} )^2}=\\\\=\frac{\sqrt{3\cdot 2} -\sqrt{3\cdot 3} -4\sqrt{2}+4\sqrt{3}}{2-3}= \frac{\sqrt{6} -\sqrt{9} -4\sqrt{2}+4\sqrt{3}}{-1}=\\\\= \frac{\sqrt{6} -3 -4\sqrt{2}+4\sqrt{3}}{-1}= -(\sqrt{6} -3 -4\sqrt{2}+4\sqrt{3})=4\sqrt{2}-\sqrt{6} -4\sqrt{3} +3[/tex]