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

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

Whakaoti mō x, y

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

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

Ngā Upane Otinga

Whakaarohia te whārite tuatahi. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

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

Tāpirihia te 1 ki te 1, ka 2.

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

Whakangāwaritia te tauraro o \frac{2}{\sqrt{2}+2} mā te whakarea i te taurunga me te tauraro ki te \sqrt{2}-2.

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

Whakaarohia te \left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Pūrua \sqrt{2}. Pūrua 2.

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

Tangohia te 4 i te 2, ka -2.

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

Me whakakore te -2 me te -2.

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

Ko te tauaro o -\left(\sqrt{2}-2\right) ko \sqrt{2}-2.

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

Tangohia te 2 i te 1, ka -1.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(\sqrt{2}+1\right)^{2}.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Tāpirihia te 2 ki te 1, ka 3.

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

Tāpirihia te 3 ki te 1, ka 4.

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

Tāpirihia te 1 ki te 1, ka 2.

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

Whakangāwaritia te tauraro o \frac{4+2\sqrt{2}-2\left(\sqrt{2}+1\right)}{\sqrt{2}+2} mā te whakarea i te taurunga me te tauraro ki te \sqrt{2}-2.

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

Whakaarohia te \left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Pūrua \sqrt{2}. Pūrua 2.

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

Tangohia te 4 i te 2, ka -2.

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

Whakawehe -1+\sqrt{2} ki te \frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2} mā te whakarea -1+\sqrt{2} ki te tau huripoki o \frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2}.

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

Whakamahia te āhuatanga tohatoha hei whakarea te -1+\sqrt{2} ki te -2.

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

Whakamahia te āhuatanga tohatoha hei whakarea te -2 ki te \sqrt{2}+1.

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

Pahekotia te 2\sqrt{2} me -2\sqrt{2}, ka 0.

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

Tangohia te 2 i te 4, ka 2.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te \sqrt{2}-2.

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

Whakangāwaritia te tauraro o \frac{2-2\sqrt{2}}{2\sqrt{2}-4} mā te whakarea i te taurunga me te tauraro ki te 2\sqrt{2}+4.

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

Whakaarohia te \left(2\sqrt{2}-4\right)\left(2\sqrt{2}+4\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Whakarohaina te \left(2\sqrt{2}\right)^{2}.

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

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Whakareatia te 4 ki te 2, ka 8.

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

Tātaihia te 4 mā te pū o 2, kia riro ko 16.

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

Tangohia te 16 i te 8, ka -8.

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

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te 2-2\sqrt{2} ki te 2\sqrt{2}+4 ka whakakotahi i ngā kupu rite.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Whakareatia te -4 ki te 2, ka -8.

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

Tangohia te 8 i te 8, ka 0.

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

Whakawehea te -4\sqrt{2} ki te -8, kia riro ko \frac{1}{2}\sqrt{2}.

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

Kua oti te pūnaha te whakatau.

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