Resolver {c}{(1-2/x+1)divx^2-2x+1/x+1=y}{x=sqrt{2}+1}

$$\left. \begin{array} { c } { ( 1 - \frac { 2 } { x + 1 } ) \div \frac { x ^ { 2 } - 2 x + 1 } { x + 1 } = y } \\ { x = \sqrt { 2 } + 1 } \end{array} \right.$$

Resolver x, y

$$x=\sqrt{2}+1\approx 2.414213562$$

$$y=\frac{\sqrt{2}}{2}\approx 0.707106781$$

Pasos da solución

Ten en conta a primeira ecuación. Insire os valores coñecidos das variables na ecuación.

$$\frac{1-\frac{2}{\sqrt{2}+1+1}}{\frac{\left(\sqrt{2}+1\right)^{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Suma 1 e 1 para obter 2.

$$\frac{1-\frac{2}{\sqrt{2}+2}}{\frac{\left(\sqrt{2}+1\right)^{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Racionaliza o denominador de \frac{2}{\sqrt{2}+2} mediante a multiplicación do numerador e o denominador por \sqrt{2}-2.

$$\frac{1-\frac{2\left(\sqrt{2}-2\right)}{\left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right)}}{\frac{\left(\sqrt{2}+1\right)^{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Considera \left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right). A multiplicación pódese transformar na diferencia de cadrados mediante a regra: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{1-\frac{2\left(\sqrt{2}-2\right)}{\left(\sqrt{2}\right)^{2}-2^{2}}}{\frac{\left(\sqrt{2}+1\right)^{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Eleva \sqrt{2} ao cadrado. Eleva 2 ao cadrado.

$$\frac{1-\frac{2\left(\sqrt{2}-2\right)}{2-4}}{\frac{\left(\sqrt{2}+1\right)^{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Resta 4 de 2 para obter -2.

$$\frac{1-\frac{2\left(\sqrt{2}-2\right)}{-2}}{\frac{\left(\sqrt{2}+1\right)^{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Anula -2 e -2.

$$\frac{1-\left(-\left(\sqrt{2}-2\right)\right)}{\frac{\left(\sqrt{2}+1\right)^{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

O contrario de -\left(\sqrt{2}-2\right) é \sqrt{2}-2.

$$\frac{1+\sqrt{2}-2}{\frac{\left(\sqrt{2}+1\right)^{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Resta 2 de 1 para obter -1.

$$\frac{-1+\sqrt{2}}{\frac{\left(\sqrt{2}+1\right)^{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Usar teorema binomial \left(a+b\right)^{2}=a^{2}+2ab+b^{2} para expandir \left(\sqrt{2}+1\right)^{2}.

$$\frac{-1+\sqrt{2}}{\frac{\left(\sqrt{2}\right)^{2}+2\sqrt{2}+1-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

O cadrado de \sqrt{2} é 2.

$$\frac{-1+\sqrt{2}}{\frac{2+2\sqrt{2}+1-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Suma 2 e 1 para obter 3.

$$\frac{-1+\sqrt{2}}{\frac{3+2\sqrt{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Suma 3 e 1 para obter 4.

$$\frac{-1+\sqrt{2}}{\frac{4+2\sqrt{2}-2\left(\sqrt{2}+1\right)}{\sqrt{2}+1+1}}=y$$

Suma 1 e 1 para obter 2.

$$\frac{-1+\sqrt{2}}{\frac{4+2\sqrt{2}-2\left(\sqrt{2}+1\right)}{\sqrt{2}+2}}=y$$

Racionaliza o denominador de \frac{4+2\sqrt{2}-2\left(\sqrt{2}+1\right)}{\sqrt{2}+2} mediante a multiplicación do numerador e o denominador por \sqrt{2}-2.

$$\frac{-1+\sqrt{2}}{\frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{\left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right)}}=y$$

Considera \left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right). A multiplicación pódese transformar na diferencia de cadrados mediante a regra: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{-1+\sqrt{2}}{\frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{\left(\sqrt{2}\right)^{2}-2^{2}}}=y$$

Eleva \sqrt{2} ao cadrado. Eleva 2 ao cadrado.

$$\frac{-1+\sqrt{2}}{\frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{2-4}}=y$$

Resta 4 de 2 para obter -2.

$$\frac{-1+\sqrt{2}}{\frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2}}=y$$

Divide -1+\sqrt{2} entre \frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2} mediante a multiplicación de -1+\sqrt{2} polo recíproco de \frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2}.

$$\frac{\left(-1+\sqrt{2}\right)\left(-2\right)}{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}=y$$

Usa a propiedade distributiva para multiplicar -1+\sqrt{2} por -2.

$$\frac{2-2\sqrt{2}}{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}=y$$

Usa a propiedade distributiva para multiplicar -2 por \sqrt{2}+1.

$$\frac{2-2\sqrt{2}}{\left(4+2\sqrt{2}-2\sqrt{2}-2\right)\left(\sqrt{2}-2\right)}=y$$

Combina 2\sqrt{2} e -2\sqrt{2} para obter 0.

$$\frac{2-2\sqrt{2}}{\left(4-2\right)\left(\sqrt{2}-2\right)}=y$$

Resta 2 de 4 para obter 2.

$$\frac{2-2\sqrt{2}}{2\left(\sqrt{2}-2\right)}=y$$

Usa a propiedade distributiva para multiplicar 2 por \sqrt{2}-2.

$$\frac{2-2\sqrt{2}}{2\sqrt{2}-4}=y$$

Racionaliza o denominador de \frac{2-2\sqrt{2}}{2\sqrt{2}-4} mediante a multiplicación do numerador e o denominador por 2\sqrt{2}+4.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{\left(2\sqrt{2}-4\right)\left(2\sqrt{2}+4\right)}=y$$

Considera \left(2\sqrt{2}-4\right)\left(2\sqrt{2}+4\right). A multiplicación pódese transformar na diferencia de cadrados mediante a regra: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{\left(2\sqrt{2}\right)^{2}-4^{2}}=y$$

Expande \left(2\sqrt{2}\right)^{2}.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{2^{2}\left(\sqrt{2}\right)^{2}-4^{2}}=y$$

Calcula 2 á potencia de 2 e obtén 4.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{4\left(\sqrt{2}\right)^{2}-4^{2}}=y$$

O cadrado de \sqrt{2} é 2.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{4\times 2-4^{2}}=y$$

Multiplica 4 e 2 para obter 8.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{8-4^{2}}=y$$

Calcula 4 á potencia de 2 e obtén 16.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{8-16}=y$$

Resta 16 de 8 para obter -8.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{-8}=y$$

Cambia de lado para que todos os termos variables estean no lado esquerdo.

$$y=\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{-8}$$

Usa a propiedade distributiva para multiplicar 2-2\sqrt{2} por 2\sqrt{2}+4 e combina os termos semellantes.

$$y=\frac{-4\sqrt{2}+8-4\left(\sqrt{2}\right)^{2}}{-8}$$

O cadrado de \sqrt{2} é 2.

$$y=\frac{-4\sqrt{2}+8-4\times 2}{-8}$$

Multiplica -4 e 2 para obter -8.

$$y=\frac{-4\sqrt{2}+8-8}{-8}$$

Resta 8 de 8 para obter 0.

$$y=\frac{-4\sqrt{2}}{-8}$$

Divide -4\sqrt{2} entre -8 para obter \frac{1}{2}\sqrt{2}.

$$y=\frac{1}{2}\sqrt{2}$$

O sistema xa funciona correctamente.

$$x=\sqrt{2}+1 y=\frac{1}{2}\sqrt{2}$$