Whakaoti i te (1/5x+1)(1-1/5x)+(x/5-5/3)^2=0

$$( \frac { 1 } { 5 } x + 1 ) ( 1 - \frac { 1 } { 5 } x ) + ( \frac { x } { 5 } - \frac { 5 } { 3 } ) ^ { 2 } = 0$$

Whakaoti mō x

$$x = \frac{17}{3} = 5\frac{2}{3} \approx 5.666666667$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakaarohia te \left(\frac{1}{5}x+1\right)\left(1-\frac{1}{5}x\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

Whakarohaina te \left(\frac{1}{5}x\right)^{2}.

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

Tātaihia te \frac{1}{5} mā te pū o 2, kia riro ko \frac{1}{25}.

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

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o 5 me 3 ko 15. Whakareatia \frac{x}{5} ki te \frac{3}{3}. Whakareatia \frac{5}{3} ki te \frac{5}{5}.

$$1-\frac{1}{25}x^{2}+\left(\frac{3x}{15}-\frac{5\times 5}{15}\right)^{2}=0$$

Tā te mea he rite te tauraro o \frac{3x}{15} me \frac{5\times 5}{15}, me tango rāua mā te tango i ō raua taurunga.

$$1-\frac{1}{25}x^{2}+\left(\frac{3x-5\times 5}{15}\right)^{2}=0$$

Mahia ngā whakarea i roto o 3x-5\times 5.

$$1-\frac{1}{25}x^{2}+\left(\frac{3x-25}{15}\right)^{2}=0$$

Kia whakarewa i te \frac{3x-25}{15} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$1-\frac{1}{25}x^{2}+\frac{\left(3x-25\right)^{2}}{15^{2}}=0$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(3x-25\right)^{2}.

$$1-\frac{1}{25}x^{2}+\frac{9x^{2}-150x+625}{15^{2}}=0$$

Tātaihia te 15 mā te pū o 2, kia riro ko 225.

$$1-\frac{1}{25}x^{2}+\frac{9x^{2}-150x+625}{225}=0$$

Whakawehea ia wā o 9x^{2}-150x+625 ki te 225, kia riro ko \frac{1}{25}x^{2}-\frac{2}{3}x+\frac{25}{9}.

$$1-\frac{1}{25}x^{2}+\frac{1}{25}x^{2}-\frac{2}{3}x+\frac{25}{9}=0$$

Pahekotia te -\frac{1}{25}x^{2} me \frac{1}{25}x^{2}, ka 0.

$$1-\frac{2}{3}x+\frac{25}{9}=0$$

Tāpirihia te 1 ki te \frac{25}{9}, ka \frac{34}{9}.

$$\frac{34}{9}-\frac{2}{3}x=0$$

Tangohia te \frac{34}{9} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-\frac{2}{3}x=-\frac{34}{9}$$

Me whakarea ngā taha e rua ki te -\frac{3}{2}, te tau utu o -\frac{2}{3}.

$$x=-\frac{34}{9}\left(-\frac{3}{2}\right)$$

Whakareatia te -\frac{34}{9} ki te -\frac{3}{2}, ka \frac{17}{3}.

$$x=\frac{17}{3}$$