Whakaoti i te sqrt{(1+6^2)[(11/3)^2-4*121/36]}

$$\sqrt { ( 1 + 6 ^ { 2 } ) [ ( \frac { 11 } { 3 } ) ^ { 2 } - 4 \times \frac { 121 } { 36 } ] }$$

Aromātai

$$0$$

Ngā Upane Otinga

Tātaihia te 6 mā te pū o 2, kia riro ko 36.

$$\sqrt{\left(1+36\right)\left(\left(\frac{11}{3}\right)^{2}-4\times \frac{121}{36}\right)}$$

Tāpirihia te 1 ki te 36, ka 37.

$$\sqrt{37\left(\left(\frac{11}{3}\right)^{2}-4\times \frac{121}{36}\right)}$$

Tātaihia te \frac{11}{3} mā te pū o 2, kia riro ko \frac{121}{9}.

$$\sqrt{37\left(\frac{121}{9}-4\times \frac{121}{36}\right)}$$

Tuhia te 4\times \frac{121}{36} hei hautanga kotahi.

$$\sqrt{37\left(\frac{121}{9}-\frac{4\times 121}{36}\right)}$$

Whakareatia te 4 ki te 121, ka 484.

$$\sqrt{37\left(\frac{121}{9}-\frac{484}{36}\right)}$$

Whakahekea te hautanga \frac{484}{36} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$\sqrt{37\left(\frac{121}{9}-\frac{121}{9}\right)}$$

Tangohia te \frac{121}{9} i te \frac{121}{9}, ka 0.

$$\sqrt{37\times 0}$$

Whakareatia te 37 ki te 0, ka 0.

$$\sqrt{0}$$

Tātaitia te pūtakerua o 0 kia tae ki 0.

$$0$$

Tauwehe

$$0$$