Whakaoti i te y=(-5)^2-4^2+(1/5)^0/3^-2+1?

$$y = \frac { ( - 5 ) ^ { 2 } - 4 ^ { 2 } + ( 1 / 5 ) ^ { 0 } } { 3 ^ { - 2 } + 1 } ?$$

Whakaoti mō y

$$y=9$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tātaihia te -5 mā te pū o 2, kia riro ko 25.

$$y=\frac{25-4^{2}+\left(\frac{1}{5}\right)^{0}}{3^{-2}+1}$$

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

$$y=\frac{25-16+\left(\frac{1}{5}\right)^{0}}{3^{-2}+1}$$

Tangohia te 16 i te 25, ka 9.

$$y=\frac{9+\left(\frac{1}{5}\right)^{0}}{3^{-2}+1}$$

Tātaihia te \frac{1}{5} mā te pū o 0, kia riro ko 1.

$$y=\frac{9+1}{3^{-2}+1}$$

Tāpirihia te 9 ki te 1, ka 10.

$$y=\frac{10}{3^{-2}+1}$$

Tātaihia te 3 mā te pū o -2, kia riro ko \frac{1}{9}.

$$y=\frac{10}{\frac{1}{9}+1}$$

Tāpirihia te \frac{1}{9} ki te 1, ka \frac{10}{9}.

$$y=\frac{10}{\frac{10}{9}}$$

Whakawehe 10 ki te \frac{10}{9} mā te whakarea 10 ki te tau huripoki o \frac{10}{9}.

$$y=10\times \frac{9}{10}$$

Whakareatia te 10 ki te \frac{9}{10}, ka 9.

$$y=9$$

Tautapa y

$$y≔9$$