Whakaoti i te 22x+1/x+1-sqrt{2}=0

$$2 \frac { 2 x + 1 } { x + 1 } - \sqrt { 2 } = 0$$

Whakaoti mō x

$$x=\frac{\sqrt{2}-3}{7}\approx -0.22654092$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe x ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x+1.

$$2\left(2x+1\right)-\sqrt{2}\left(x+1\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te 2x+1.

$$4x+2-\sqrt{2}\left(x+1\right)=0$$

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

$$4x+2-\sqrt{2}x-\sqrt{2}=0$$

Tangohia te 2 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$4x-\sqrt{2}x-\sqrt{2}=-2$$

Me tāpiri te \sqrt{2} ki ngā taha e rua.

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

Pahekotia ngā kīanga tau katoa e whai ana i te x.

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te 4-\sqrt{2}.

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

Mā te whakawehe ki te 4-\sqrt{2} ka wetekia te whakareanga ki te 4-\sqrt{2}.

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

Whakawehe -2+\sqrt{2} ki te 4-\sqrt{2}.

$$x=\frac{\sqrt{2}-3}{7}$$