Whakaoti i te sqrt{2x+1}-sqrt{x+2}=2sqrt{3}

$$\sqrt { 2 x + 1 } - \sqrt { x + 2 } = 2 \sqrt { 3 }$$

Whakaoti mō x

$$x=73$$

Ngā Upane Otinga

Me tango -\sqrt{x+2} mai i ngā taha e rua o te whārite.

$$\sqrt{2x+1}=2\sqrt{3}+\sqrt{x+2}$$

Pūruatia ngā taha e rua o te whārite.

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

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

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2\sqrt{3}+\sqrt{x+2}\right)^{2}.

$$2x+1=4\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$2x+1=4\times 3+4\sqrt{3}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}$$

Whakareatia te 4 ki te 3, ka 12.

$$2x+1=12+4\sqrt{3}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}$$

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

$$2x+1=12+4\sqrt{3}\sqrt{x+2}+x+2$$

Tāpirihia te 12 ki te 2, ka 14.

$$2x+1=14+4\sqrt{3}\sqrt{x+2}+x$$

Me tango 14+x mai i ngā taha e rua o te whārite.

$$2x+1-\left(14+x\right)=4\sqrt{3}\sqrt{x+2}$$

Hei kimi i te tauaro o 14+x, kimihia te tauaro o ia taurangi.

$$2x+1-14-x=4\sqrt{3}\sqrt{x+2}$$

Tangohia te 14 i te 1, ka -13.

$$2x-13-x=4\sqrt{3}\sqrt{x+2}$$

Pahekotia te 2x me -x, ka x.

$$x-13=4\sqrt{3}\sqrt{x+2}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(x-13\right)^{2}=\left(4\sqrt{3}\sqrt{x+2}\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-13\right)^{2}.

$$x^{2}-26x+169=\left(4\sqrt{3}\sqrt{x+2}\right)^{2}$$

Whakarohaina te \left(4\sqrt{3}\sqrt{x+2}\right)^{2}.

$$x^{2}-26x+169=4^{2}\left(\sqrt{3}\right)^{2}\left(\sqrt{x+2}\right)^{2}$$

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

$$x^{2}-26x+169=16\left(\sqrt{3}\right)^{2}\left(\sqrt{x+2}\right)^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$x^{2}-26x+169=16\times 3\left(\sqrt{x+2}\right)^{2}$$

Whakareatia te 16 ki te 3, ka 48.

$$x^{2}-26x+169=48\left(\sqrt{x+2}\right)^{2}$$

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

$$x^{2}-26x+169=48\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 48 ki te x+2.

$$x^{2}-26x+169=48x+96$$

Tangohia te 48x mai i ngā taha e rua.

$$x^{2}-26x+169-48x=96$$

Pahekotia te -26x me -48x, ka -74x.

$$x^{2}-74x+169=96$$

Tangohia te 96 mai i ngā taha e rua.

$$x^{2}-74x+169-96=0$$

Tangohia te 96 i te 169, ka 73.

$$x^{2}-74x+73=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-74x+73 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-74 ab=73$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-73 b=-1$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

$$\left(x-73\right)\left(x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-73=0 me te x-1=0.

$$x=73 x=1$$

Whakakapia te 73 mō te x i te whārite \sqrt{2x+1}-\sqrt{x+2}=2\sqrt{3}.

$$\sqrt{2\times 73+1}-\sqrt{73+2}=2\sqrt{3}$$

Whakarūnātia. Ko te uara x=73 kua ngata te whārite.

$$2\times 3^{\frac{1}{2}}=2\times 3^{\frac{1}{2}}$$

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

$$\sqrt{2\times 1+1}-\sqrt{1+2}=2\sqrt{3}$$

Whakarūnātia. Ko te uara x=1 kāore e ngata ana ki te whārite.

$$0=2\times 3^{\frac{1}{2}}$$

Whakakapia te 73 mō te x i te whārite \sqrt{2x+1}-\sqrt{x+2}=2\sqrt{3}.

$$\sqrt{2\times 73+1}-\sqrt{73+2}=2\sqrt{3}$$

Whakarūnātia. Ko te uara x=73 kua ngata te whārite.

$$2\times 3^{\frac{1}{2}}=2\times 3^{\frac{1}{2}}$$

Ko te whārite \sqrt{2x+1}=\sqrt{x+2}+2\sqrt{3} he rongoā ahurei.

$$x=73$$