Question

Ayudenme ;-;

A) 2/(3+ √(2 )) - 1/√2=

B) 2x+5=(7x^(2 )-2)/(x+1)=

Answer 1

Respuesta:

Explicación paso a paso:

$\frac{2}{3+\sqrt{2}}-\frac{1}{\sqrt{2}}=\frac{(\sqrt{2})(2)-(3+\sqrt{2})(1)}{(3+\sqrt{2})(\sqrt{2})}=\\\\=\frac{2\sqrt{2}-3-\sqrt{2}}{3\sqrt{2}+(\sqrt{2})^{2}}=\frac{\sqrt{2}-3}{3+\sqrt{2}+2}=\frac{\sqrt{2}-3}{\sqrt{2}+5}\\\\2x+5=\frac{7x^{2}-2}{x+1}$

$\texttt{Multiplicando por x+1 la ecuaci\'on nos queda:}\\\\(x+1)(2x+5)=(x+1)*\frac{7x^{2}-2}{x+1}\\\\(x+1)(2x+5)=7x^{2}-2\\\\\texttt{Transponiendo y simplificando:}\\(x+1)(2x+5)-7x^{2}+2=0\\2x+5+2x^{2}+5x-7x^{2}+2=0\\2x^{2}-7x^{2}+2x+5x+5+2=0\\-5x^{2}+7x+7=0$

$\texttt{Resolviendo la \'ultima expresi\'on, que es una}\\\texttt{ecuaci\'on cuadr\'atica,nos queda que:}$

$-5x^{2}+7x+7=0\\x=\frac{-7\pm\sqrt{7^{2}-4(-5)(7)}}{2(-5)}\\x=\frac{-7\pm\sqrt{49+140}}{-10}\\x=\frac{-7\pm\sqrt{189}}{-10}\\x=\frac{-7\pm3\sqrt{21}}{-10}\\x_{1}=\frac{7+3\sqrt{21}}{10}\\x_{2}=\frac{7-3\sqrt{21}}{10}$

Answer 2

Respuesta:

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