Whakaoti i te {l}{C=sqrt{8}+sqrt{8}}{D=sqrt{8}-sqrt{8}}{a=C*D}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { C = \sqrt{8} + \sqrt{8} }\\ { D = \sqrt{8} - \sqrt{8} }\\ { a = C \cdot D }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Whakaoti mō C, D, a, b, c, d

$$d=0$$

Ngā Upane Otinga

Whakaarohia te whārite tuatahi. Tauwehea te 8=2^{2}\times 2. Tuhia anō te pūtake rua o te hua \sqrt{2^{2}\times 2} hei hua o ngā pūtake rua \sqrt{2^{2}}\sqrt{2}. Tuhia te pūtakerua o te 2^{2}.

$$C=2\sqrt{2}+\sqrt{8}$$

Tauwehea te 8=2^{2}\times 2. Tuhia anō te pūtake rua o te hua \sqrt{2^{2}\times 2} hei hua o ngā pūtake rua \sqrt{2^{2}}\sqrt{2}. Tuhia te pūtakerua o te 2^{2}.

$$C=2\sqrt{2}+2\sqrt{2}$$

Pahekotia te 2\sqrt{2} me 2\sqrt{2}, ka 4\sqrt{2}.

$$C=4\sqrt{2}$$

Whakaarohia te whārite tuarua. Tauwehea te 8=2^{2}\times 2. Tuhia anō te pūtake rua o te hua \sqrt{2^{2}\times 2} hei hua o ngā pūtake rua \sqrt{2^{2}}\sqrt{2}. Tuhia te pūtakerua o te 2^{2}.

$$D=2\sqrt{2}-\sqrt{8}$$

Tauwehea te 8=2^{2}\times 2. Tuhia anō te pūtake rua o te hua \sqrt{2^{2}\times 2} hei hua o ngā pūtake rua \sqrt{2^{2}}\sqrt{2}. Tuhia te pūtakerua o te 2^{2}.

$$D=2\sqrt{2}-2\sqrt{2}$$

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

$$D=0$$

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

$$a=4\sqrt{2}\times 0$$

Whakareatia te 4 ki te 0, ka 0.

$$a=0\sqrt{2}$$

Ko te tau i whakarea ki te kore ka hua ko te kore.

$$a=0$$

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

$$b=0$$

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

$$c=0$$

Whakaarohia te whārite (6). Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$d=0$$

Kua oti te pūnaha te whakatau.

$$C=4\sqrt{2} D=0 a=0 b=0 c=0 d=0$$