Whakaoti i te 5x^2=4x

$$5 x ^ { 2 } = 4 x$$

Whakaoti mō x

$$x=\frac{4}{5}=0.8$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 4x mai i ngā taha e rua.

$$5x^{2}-4x=0$$

Tauwehea te x.

$$x\left(5x-4\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 5x-4=0.

$$x=0 x=\frac{4}{5}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 4x mai i ngā taha e rua.

$$5x^{2}-4x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -4 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times 5}$$

Tuhia te pūtakerua o te \left(-4\right)^{2}.

$$x=\frac{-\left(-4\right)±4}{2\times 5}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±4}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{4±4}{10}$$

Nā, me whakaoti te whārite x=\frac{4±4}{10} ina he tāpiri te ±. Tāpiri 4 ki te 4.

$$x=\frac{8}{10}$$

Whakahekea te hautanga \frac{8}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{4}{5}$$

Nā, me whakaoti te whārite x=\frac{4±4}{10} ina he tango te ±. Tango 4 mai i 4.

$$x=\frac{0}{10}$$

Whakawehe 0 ki te 10.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=\frac{4}{5} x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 4x mai i ngā taha e rua.

$$5x^{2}-4x=0$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}-4x}{5}=\frac{0}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

$$x^{2}-\frac{4}{5}x=\frac{0}{5}$$

Whakawehe 0 ki te 5.

$$x^{2}-\frac{4}{5}x=0$$

Whakawehea te -\frac{4}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{2}{5}. Nā, tāpiria te pūrua o te -\frac{2}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=\left(-\frac{2}{5}\right)^{2}$$

Pūruatia -\frac{2}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{4}{25}$$

Tauwehea x^{2}-\frac{4}{5}x+\frac{4}{25}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{4}{25}$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Whakarūnātia.

$$x-\frac{2}{5}=\frac{2}{5} x-\frac{2}{5}=-\frac{2}{5}$$

Me tāpiri \frac{2}{5} ki ngā taha e rua o te whārite.

$$x=\frac{4}{5} x=0$$