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

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

Réitigh do x,y.

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

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

Céimeanna Réitigh

Cuir an chéad cothromóid san áireamh. Ionsáigh luachanna aitheanta na n-athróg sa chothromóid.

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

Suimigh 1 agus 1 chun 2 a fháil.

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

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi \sqrt{2}-2 chun ainmneoir \frac{2}{\sqrt{2}+2} a thiontú in uimhir chóimheasta.

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

Mar shampla \left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right). Is féidir iolrúchán a athrú ó bhonn go dtí difríocht na gcearnóg ag úsáid na rialach seo: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Cearnóg \sqrt{2}. Cearnóg 2.

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

Dealaigh 4 ó 2 chun -2 a fháil.

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

Cealaigh -2 agus -2.

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

Tá \sqrt{2}-2 urchomhairleach le -\left(\sqrt{2}-2\right).

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

Dealaigh 2 ó 1 chun -1 a fháil.

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

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(\sqrt{2}+1\right)^{2} a leathnú.

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

Is é 2 uimhir chearnach \sqrt{2}.

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

Suimigh 2 agus 1 chun 3 a fháil.

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

Suimigh 3 agus 1 chun 4 a fháil.

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

Suimigh 1 agus 1 chun 2 a fháil.

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

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi \sqrt{2}-2 chun ainmneoir \frac{4+2\sqrt{2}-2\left(\sqrt{2}+1\right)}{\sqrt{2}+2} a thiontú in uimhir chóimheasta.

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

Mar shampla \left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right). Is féidir iolrúchán a athrú ó bhonn go dtí difríocht na gcearnóg ag úsáid na rialach seo: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Cearnóg \sqrt{2}. Cearnóg 2.

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

Dealaigh 4 ó 2 chun -2 a fháil.

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

Roinn -1+\sqrt{2} faoi \frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2} trí -1+\sqrt{2} a mhéadú faoi dheilín \frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2}.

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

Úsáid an t-airí dáileach chun -1+\sqrt{2} a mhéadú faoi -2.

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

Úsáid an t-airí dáileach chun -2 a mhéadú faoi \sqrt{2}+1.

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

Comhcheangail 2\sqrt{2} agus -2\sqrt{2} chun 0 a fháil.

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

Dealaigh 2 ó 4 chun 2 a fháil.

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

Úsáid an t-airí dáileach chun 2 a mhéadú faoi \sqrt{2}-2.

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

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi 2\sqrt{2}+4 chun ainmneoir \frac{2-2\sqrt{2}}{2\sqrt{2}-4} a thiontú in uimhir chóimheasta.

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

Mar shampla \left(2\sqrt{2}-4\right)\left(2\sqrt{2}+4\right). Is féidir iolrúchán a athrú ó bhonn go dtí difríocht na gcearnóg ag úsáid na rialach seo: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Fairsingigh \left(2\sqrt{2}\right)^{2}

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

Ríomh cumhacht 2 de 2 agus faigh 4.

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

Is é 2 uimhir chearnach \sqrt{2}.

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

Méadaigh 4 agus 2 chun 8 a fháil.

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

Ríomh cumhacht 4 de 2 agus faigh 16.

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

Dealaigh 16 ó 8 chun -8 a fháil.

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

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

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

Úsáid an t-airí dáileach chun 2-2\sqrt{2} a mhéadú faoi 2\sqrt{2}+4 agus chun téarmaí comhchosúla a chumasc.

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

Is é 2 uimhir chearnach \sqrt{2}.

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

Méadaigh -4 agus 2 chun -8 a fháil.

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

Dealaigh 8 ó 8 chun 0 a fháil.

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

Roinn -4\sqrt{2} faoi -8 chun \frac{1}{2}\sqrt{2} a fháil.

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

Tá an córas réitithe anois.

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