Respuestas
Respuesta dada por:
0
4.- Racionalizar el Denominador.
![\frac{1}{ \sqrt[5]{ x^{27} y^{3} z^{11}}}
\frac{1}{ \sqrt[5]{ x^{27} y^{3} z^{11}}}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B+%5Csqrt%5B5%5D%7B+x%5E%7B27%7D+y%5E%7B3%7D+z%5E%7B11%7D%7D%7D+%0A)
![( \frac{1}{ \sqrt[5]{ x^{27} y^{3} z^{11}}}) \frac{\sqrt[5]{ x^{3} y^{2} z^{4}}}{\sqrt[5]{ x^{3} y^{2} z^{4}}}
( \frac{1}{ \sqrt[5]{ x^{27} y^{3} z^{11}}}) \frac{\sqrt[5]{ x^{3} y^{2} z^{4}}}{\sqrt[5]{ x^{3} y^{2} z^{4}}}](https://tex.z-dn.net/?f=%28+%5Cfrac%7B1%7D%7B+%5Csqrt%5B5%5D%7B+x%5E%7B27%7D+y%5E%7B3%7D+z%5E%7B11%7D%7D%7D%29+%5Cfrac%7B%5Csqrt%5B5%5D%7B+x%5E%7B3%7D+y%5E%7B2%7D+z%5E%7B4%7D%7D%7D%7B%5Csqrt%5B5%5D%7B+x%5E%7B3%7D+y%5E%7B2%7D+z%5E%7B4%7D%7D%7D+++%0A)
![\frac{\sqrt[5]{ x^{3} y^{2} z^{4}}}{ \sqrt[5]{ x^{30} y^{5} z^{15}}} \frac{\sqrt[5]{ x^{3} y^{2} z^{4}}}{ \sqrt[5]{ x^{30} y^{5} z^{15}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B5%5D%7B+x%5E%7B3%7D+y%5E%7B2%7D+z%5E%7B4%7D%7D%7D%7B+%5Csqrt%5B5%5D%7B+x%5E%7B30%7D+y%5E%7B5%7D+z%5E%7B15%7D%7D%7D+++)
![\frac{\sqrt[5]{ x^{3} y^{2} z^{4}}}{x^{6} y z^{3}} \frac{\sqrt[5]{ x^{3} y^{2} z^{4}}}{x^{6} y z^{3}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B5%5D%7B+x%5E%7B3%7D+y%5E%7B2%7D+z%5E%7B4%7D%7D%7D%7Bx%5E%7B6%7D+y+z%5E%7B3%7D%7D+++)
El denominador racionalizado queda:
La respuesta correcta es la (A)
6.- Efectua:
![\frac{2+ \sqrt{3} }{\sqrt{3}+1}+\frac{2-\sqrt{3} }{\sqrt{3}-1}=\frac{(2+\sqrt{3})(\sqrt{3}-1)+(2-\sqrt{3})(\sqrt{3}+1)}{ ( \sqrt{3} )^{2} -1} \frac{2+ \sqrt{3} }{\sqrt{3}+1}+\frac{2-\sqrt{3} }{\sqrt{3}-1}=\frac{(2+\sqrt{3})(\sqrt{3}-1)+(2-\sqrt{3})(\sqrt{3}+1)}{ ( \sqrt{3} )^{2} -1}](https://tex.z-dn.net/?f=%5Cfrac%7B2%2B+%5Csqrt%7B3%7D+%7D%7B%5Csqrt%7B3%7D%2B1%7D%2B%5Cfrac%7B2-%5Csqrt%7B3%7D+%7D%7B%5Csqrt%7B3%7D-1%7D%3D%5Cfrac%7B%282%2B%5Csqrt%7B3%7D%29%28%5Csqrt%7B3%7D-1%29%2B%282-%5Csqrt%7B3%7D%29%28%5Csqrt%7B3%7D%2B1%29%7D%7B+%28+%5Csqrt%7B3%7D+%29%5E%7B2%7D+-1%7D+)
![\frac{(2+\sqrt{3})(\sqrt{3}-1)+(2-\sqrt{3})(\sqrt{3}+1)}{3-1} =\frac{(2\sqrt{3}-2+3-\sqrt{3})+(2\sqrt{3}+2-3-\sqrt{3})}{2} \frac{(2+\sqrt{3})(\sqrt{3}-1)+(2-\sqrt{3})(\sqrt{3}+1)}{3-1} =\frac{(2\sqrt{3}-2+3-\sqrt{3})+(2\sqrt{3}+2-3-\sqrt{3})}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%282%2B%5Csqrt%7B3%7D%29%28%5Csqrt%7B3%7D-1%29%2B%282-%5Csqrt%7B3%7D%29%28%5Csqrt%7B3%7D%2B1%29%7D%7B3-1%7D+%3D%5Cfrac%7B%282%5Csqrt%7B3%7D-2%2B3-%5Csqrt%7B3%7D%29%2B%282%5Csqrt%7B3%7D%2B2-3-%5Csqrt%7B3%7D%29%7D%7B2%7D+)
![\frac{(2\sqrt{3}-2+3-\sqrt{3})+(2\sqrt{3}+2-3-\sqrt{3})}{2} =\frac{(2\sqrt{3})}{2}=\sqrt{3} \frac{(2\sqrt{3}-2+3-\sqrt{3})+(2\sqrt{3}+2-3-\sqrt{3})}{2} =\frac{(2\sqrt{3})}{2}=\sqrt{3}](https://tex.z-dn.net/?f=%5Cfrac%7B%282%5Csqrt%7B3%7D-2%2B3-%5Csqrt%7B3%7D%29%2B%282%5Csqrt%7B3%7D%2B2-3-%5Csqrt%7B3%7D%29%7D%7B2%7D+%3D%5Cfrac%7B%282%5Csqrt%7B3%7D%29%7D%7B2%7D%3D%5Csqrt%7B3%7D)
La respuesta correcta es la (D)
7.- Racionaliza e indica denominador en expresión racionalizada.
![\frac{ \sqrt{12+2 \sqrt{27} } }{ \sqrt{12-2 \sqrt{27} } } \frac{ \sqrt{12+2 \sqrt{27} } }{ \sqrt{12-2 \sqrt{27} } }](https://tex.z-dn.net/?f=+%5Cfrac%7B+%5Csqrt%7B12%2B2+%5Csqrt%7B27%7D+%7D+%7D%7B+%5Csqrt%7B12-2+%5Csqrt%7B27%7D+%7D+%7D+)
![( \frac{ \sqrt{12+2 \sqrt{27} } }{ \sqrt{12-2 \sqrt{27} } } )( \frac{ \sqrt{12+2 \sqrt{27} } }{ \sqrt{12+2 \sqrt{27} } } )=\frac{12+2 \sqrt{27 }} { \sqrt{ 12^{2} +(2 \sqrt{27})^{2} } }
( \frac{ \sqrt{12+2 \sqrt{27} } }{ \sqrt{12-2 \sqrt{27} } } )( \frac{ \sqrt{12+2 \sqrt{27} } }{ \sqrt{12+2 \sqrt{27} } } )=\frac{12+2 \sqrt{27 }} { \sqrt{ 12^{2} +(2 \sqrt{27})^{2} } }](https://tex.z-dn.net/?f=%28+%5Cfrac%7B+%5Csqrt%7B12%2B2+%5Csqrt%7B27%7D+%7D+%7D%7B+%5Csqrt%7B12-2+%5Csqrt%7B27%7D+%7D+%7D+%29%28+%5Cfrac%7B+%5Csqrt%7B12%2B2+%5Csqrt%7B27%7D+%7D+%7D%7B+%5Csqrt%7B12%2B2+%5Csqrt%7B27%7D+%7D+%7D+%29%3D%5Cfrac%7B12%2B2+%5Csqrt%7B27+%7D%7D+%7B+%5Csqrt%7B+12%5E%7B2%7D+%2B%282+%5Csqrt%7B27%7D%29%5E%7B2%7D+%7D+%7D%0A%0A)
![\frac{12+2 \sqrt{27}}{\sqrt{ 12^{2} +(2 \sqrt{27})^{2}}}=\frac{12+2 \sqrt{27} } { \sqrt{ 144 -(4)(27)} }=\frac{ 12+2 \sqrt{27} } { \sqrt{ 144 -108} } }=\frac{12+2 \sqrt{27}} { \sqrt{ 36} } }
\frac{12+2 \sqrt{27}}{\sqrt{ 12^{2} +(2 \sqrt{27})^{2}}}=\frac{12+2 \sqrt{27} } { \sqrt{ 144 -(4)(27)} }=\frac{ 12+2 \sqrt{27} } { \sqrt{ 144 -108} } }=\frac{12+2 \sqrt{27}} { \sqrt{ 36} } }](https://tex.z-dn.net/?f=%5Cfrac%7B12%2B2+%5Csqrt%7B27%7D%7D%7B%5Csqrt%7B+12%5E%7B2%7D+%2B%282+%5Csqrt%7B27%7D%29%5E%7B2%7D%7D%7D%3D%5Cfrac%7B12%2B2+%5Csqrt%7B27%7D+%7D+%7B+%5Csqrt%7B+144+-%284%29%2827%29%7D+%7D%3D%5Cfrac%7B+12%2B2+%5Csqrt%7B27%7D+%7D+%7B+%5Csqrt%7B+144+-108%7D+%7D+%7D%3D%5Cfrac%7B12%2B2+%5Csqrt%7B27%7D%7D+%7B+%5Csqrt%7B+36%7D+%7D+%7D%0A)
![\frac{12+2 \sqrt{(9)(3)}} { \sqrt{ 36} } }=\frac{12+(2)(3) \sqrt{3}} {6}=\frac{12+6 \sqrt{3}} {6}=2+\sqrt{3}
\frac{12+2 \sqrt{(9)(3)}} { \sqrt{ 36} } }=\frac{12+(2)(3) \sqrt{3}} {6}=\frac{12+6 \sqrt{3}} {6}=2+\sqrt{3}](https://tex.z-dn.net/?f=%5Cfrac%7B12%2B2+%5Csqrt%7B%289%29%283%29%7D%7D+%7B+%5Csqrt%7B+36%7D+%7D+%7D%3D%5Cfrac%7B12%2B%282%29%283%29+%5Csqrt%7B3%7D%7D+%7B6%7D%3D%5Cfrac%7B12%2B6+%5Csqrt%7B3%7D%7D+%7B6%7D%3D2%2B%5Csqrt%7B3%7D%0A%0A)
La respuesta correcta es la D
El denominador racionalizado queda:
La respuesta correcta es la (A)
6.- Efectua:
La respuesta correcta es la (D)
7.- Racionaliza e indica denominador en expresión racionalizada.
La respuesta correcta es la D
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