Question

Ayuda en la 9 y 7 con procedimiento porfa

Answer 1

Respuesta:

$(x + y)( {x}^{2} + {y}^{2} )( {x}^{4} + {y}^{4} )...( {x}^{n} + {y}^{n} ) = \frac{ {x}^{2n} - {y}^{2n} }{x - y}$

7. Hallar

$\sqrt[32]{5( {3}^{2} + {2}^{2} )( {3}^{4} + {2}^{4})( {3}^{8} + {2}^{8} ) + {2}^{16} }$

$\sqrt[32]{(3 + 2)( {3}^{2} + {2}^{2} )( {3}^{4} + {2}^{4})( {3}^{8} + {2}^{8} ) + {2}^{16} }$

$\sqrt[32]{ \frac{ {3}^{2(8)} - {2}^{2(8)} }{3 - 2} + {2}^{16} }$

$\sqrt[32]{ \frac{ {3}^{16} - {2}^{16} }{1} + {2}^{16} }$

$\sqrt[32]{ {3}^{16} - {2}^{16} + {2}^{16} }$

$\sqrt[32]{ {3}^{16}}$

${3}^{ \frac{16}{32} }$

${3}^{ \frac{1}{2} }$

$\sqrt{3}$

9. Si x+y= 23 ; xy = 1 ; calcular √x + √y

${( \sqrt{x} + \sqrt{y} )}^{2} = { \sqrt{x} }^{2} + 2 \sqrt{x} \sqrt{y} + { \sqrt{y} }^{2}$

${( \sqrt{x} + \sqrt{y} )}^{2} =x + 2 \sqrt{xy} + y$

${( \sqrt{x} + \sqrt{y} )}^{2} =x + y + 2 \sqrt{xy}$

${( \sqrt{x} + \sqrt{y} )}^{2} =23+ 2 \sqrt{1}$

${( \sqrt{x} + \sqrt{y} )}^{2} =23+ 2 (1)$

${( \sqrt{x} + \sqrt{y} )}^{2} =23+ 2$

${( \sqrt{x} + \sqrt{y} )}^{2} =25$

$\sqrt{x} + \sqrt{y} = \sqrt{25}$

$\sqrt{x} + \sqrt{y} = 5$

$11. \: \: \: Si\: \: x-\frac{1}{x}=2 \: \: \: \: \: Hallar \: \: x^2 + x^{-2}$

$x-\frac{1}{x}=2$

$(x-\frac{1}{x})^2=2^2$

$(x)^2 -2(x)(\frac{1}{x})+( \frac{1}{x})^2=4$

$x^2 -2(\frac{x}{x})+\frac{1}{x^2}=4$

$x^2 -2(1)+\frac{1}{x^2}=4$

$x^2 -2+\frac{1}{x^2}=4$

$x^2+\frac{1}{x^2}=4+2$

$x^2+\frac{1}{x^2}=6$

$x^2+x^{-2}=6$