Whakaoti i te sqrt{x^2-1}=sqrt{2x+1}

$$\sqrt { x ^ { 2 } - 1 } = \sqrt { 2 x + 1 }$$

Whakaoti mō x (complex solution)

$$x=\sqrt{3}+1\approx 2.732050808$$

$$x=1-\sqrt{3}\approx -0.732050808$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Tātaihia te \sqrt{x^{2}-1} mā te pū o 2, kia riro ko x^{2}-1.

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

Tātaihia te \sqrt{2x+1} mā te pū o 2, kia riro ko 2x+1.

$$x^{2}-1=2x+1$$

Tangohia te 2x mai i ngā taha e rua.

$$x^{2}-1-2x=1$$

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}-1-2x-1=0$$

Tangohia te 1 i te -1, ka -2.

$$x^{2}-2-2x=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x^{2}-2x-2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -2 mō b, me -2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)}}{2}$$

Pūrua -2.

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

Whakareatia -4 ki te -2.

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

Tāpiri 4 ki te 8.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

Tuhia te pūtakerua o te 12.

$$x=\frac{-\left(-2\right)±2\sqrt{3}}{2}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{3}}{2}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{3}}{2} ina he tāpiri te ±. Tāpiri 2 ki te 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+2}{2}$$

Whakawehe 2+2\sqrt{3} ki te 2.

$$x=\sqrt{3}+1$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{3}}{2} ina he tango te ±. Tango 2\sqrt{3} mai i 2.

$$x=\frac{2-2\sqrt{3}}{2}$$

Whakawehe 2-2\sqrt{3} ki te 2.

$$x=1-\sqrt{3}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$

Whakakapia te \sqrt{3}+1 mō te x i te whārite \sqrt{x^{2}-1}=\sqrt{2x+1}.

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

Whakarūnātia. Ko te uara x=\sqrt{3}+1 kua ngata te whārite.

$$\left(3+2\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(2\times 3^{\frac{1}{2}}+3\right)^{\frac{1}{2}}$$

Whakakapia te 1-\sqrt{3} mō te x i te whārite \sqrt{x^{2}-1}=\sqrt{2x+1}.

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

Whakarūnātia. Ko te uara x=1-\sqrt{3} kua ngata te whārite.

$$i\left(-\left(3-2\times 3^{\frac{1}{2}}\right)\right)^{\frac{1}{2}}=i\left(-\left(3-2\times 3^{\frac{1}{2}}\right)\right)^{\frac{1}{2}}$$

Rārangihia ngā rongoā katoa o \sqrt{x^{2}-1}=\sqrt{2x+1}.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$

Whakaoti mō x

$$x=\sqrt{3}+1\approx 2.732050808$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Tātaihia te \sqrt{x^{2}-1} mā te pū o 2, kia riro ko x^{2}-1.

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

Tātaihia te \sqrt{2x+1} mā te pū o 2, kia riro ko 2x+1.

$$x^{2}-1=2x+1$$

Tangohia te 2x mai i ngā taha e rua.

$$x^{2}-1-2x=1$$

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}-1-2x-1=0$$

Tangohia te 1 i te -1, ka -2.

$$x^{2}-2-2x=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x^{2}-2x-2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -2 mō b, me -2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)}}{2}$$

Pūrua -2.

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

Whakareatia -4 ki te -2.

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

Tāpiri 4 ki te 8.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

Tuhia te pūtakerua o te 12.

$$x=\frac{-\left(-2\right)±2\sqrt{3}}{2}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{3}}{2}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{3}}{2} ina he tāpiri te ±. Tāpiri 2 ki te 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+2}{2}$$

Whakawehe 2+2\sqrt{3} ki te 2.

$$x=\sqrt{3}+1$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{3}}{2} ina he tango te ±. Tango 2\sqrt{3} mai i 2.

$$x=\frac{2-2\sqrt{3}}{2}$$

Whakawehe 2-2\sqrt{3} ki te 2.

$$x=1-\sqrt{3}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$

Whakakapia te \sqrt{3}+1 mō te x i te whārite \sqrt{x^{2}-1}=\sqrt{2x+1}.

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

Whakarūnātia. Ko te uara x=\sqrt{3}+1 kua ngata te whārite.

$$\left(3+2\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(2\times 3^{\frac{1}{2}}+3\right)^{\frac{1}{2}}$$

Whakakapia te 1-\sqrt{3} mō te x i te whārite \sqrt{x^{2}-1}=\sqrt{2x+1}. Te kīanga \sqrt{\left(1-\sqrt{3}\right)^{2}-1} kia kore e tautuhitia nā te mea kāore te radicand e noho tōraro.

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

Ko te whārite \sqrt{x^{2}-1}=\sqrt{2x+1} he rongoā ahurei.

$$x=\sqrt{3}+1$$