Whakaoti i te (5x-2)^2-(2x-1)(2x+1)=47+x

$$( 5 x - 2 ) ^ { 2 } - ( 2 x - 1 ) ( 2 x + 1 ) = 47 + x$$

Whakaoti mō x

$$x=-1$$

$$x=2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(5x-2\right)^{2}.

$$25x^{2}-20x+4-\left(2x-1\right)\left(2x+1\right)=47+x$$

Whakaarohia te \left(2x-1\right)\left(2x+1\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.

$$25x^{2}-20x+4-\left(\left(2x\right)^{2}-1\right)=47+x$$

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

$$25x^{2}-20x+4-\left(2^{2}x^{2}-1\right)=47+x$$

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

$$25x^{2}-20x+4-\left(4x^{2}-1\right)=47+x$$

Hei kimi i te tauaro o 4x^{2}-1, kimihia te tauaro o ia taurangi.

$$25x^{2}-20x+4-4x^{2}+1=47+x$$

Pahekotia te 25x^{2} me -4x^{2}, ka 21x^{2}.

$$21x^{2}-20x+4+1=47+x$$

Tāpirihia te 4 ki te 1, ka 5.

$$21x^{2}-20x+5=47+x$$

Tangohia te 47 mai i ngā taha e rua.

$$21x^{2}-20x+5-47=x$$

Tangohia te 47 i te 5, ka -42.

$$21x^{2}-20x-42=x$$

Tangohia te x mai i ngā taha e rua.

$$21x^{2}-20x-42-x=0$$

Pahekotia te -20x me -x, ka -21x.

$$21x^{2}-21x-42=0$$

Whakawehea ngā taha e rua ki te 21.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=1\left(-2\right)=-2$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=1$$

Tuhia anō te x^{2}-x-2 hei \left(x^{2}-2x\right)+\left(x-2\right).

$$\left(x^{2}-2x\right)+\left(x-2\right)$$

Whakatauwehea atu x i te x^{2}-2x.

$$x\left(x-2\right)+x-2$$

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-2=0 me te x+1=0.

$$x=2 x=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(5x-2\right)^{2}.

$$25x^{2}-20x+4-\left(2x-1\right)\left(2x+1\right)=47+x$$

Whakaarohia te \left(2x-1\right)\left(2x+1\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.

$$25x^{2}-20x+4-\left(\left(2x\right)^{2}-1\right)=47+x$$

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

$$25x^{2}-20x+4-\left(2^{2}x^{2}-1\right)=47+x$$

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

$$25x^{2}-20x+4-\left(4x^{2}-1\right)=47+x$$

Hei kimi i te tauaro o 4x^{2}-1, kimihia te tauaro o ia taurangi.

$$25x^{2}-20x+4-4x^{2}+1=47+x$$

Pahekotia te 25x^{2} me -4x^{2}, ka 21x^{2}.

$$21x^{2}-20x+4+1=47+x$$

Tāpirihia te 4 ki te 1, ka 5.

$$21x^{2}-20x+5=47+x$$

Tangohia te 47 mai i ngā taha e rua.

$$21x^{2}-20x+5-47=x$$

Tangohia te 47 i te 5, ka -42.

$$21x^{2}-20x-42=x$$

Tangohia te x mai i ngā taha e rua.

$$21x^{2}-20x-42-x=0$$

Pahekotia te -20x me -x, ka -21x.

$$21x^{2}-21x-42=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 21 mō a, -21 mō b, me -42 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 21\left(-42\right)}}{2\times 21}$$

Pūrua -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 21\left(-42\right)}}{2\times 21}$$

Whakareatia -4 ki te 21.

$$x=\frac{-\left(-21\right)±\sqrt{441-84\left(-42\right)}}{2\times 21}$$

Whakareatia -84 ki te -42.

$$x=\frac{-\left(-21\right)±\sqrt{441+3528}}{2\times 21}$$

Tāpiri 441 ki te 3528.

$$x=\frac{-\left(-21\right)±\sqrt{3969}}{2\times 21}$$

Tuhia te pūtakerua o te 3969.

$$x=\frac{-\left(-21\right)±63}{2\times 21}$$

Ko te tauaro o -21 ko 21.

$$x=\frac{21±63}{2\times 21}$$

Whakareatia 2 ki te 21.

$$x=\frac{21±63}{42}$$

Nā, me whakaoti te whārite x=\frac{21±63}{42} ina he tāpiri te ±. Tāpiri 21 ki te 63.

$$x=\frac{84}{42}$$

Whakawehe 84 ki te 42.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{21±63}{42} ina he tango te ±. Tango 63 mai i 21.

$$x=-\frac{42}{42}$$

Whakawehe -42 ki te 42.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=2 x=-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(5x-2\right)^{2}.

$$25x^{2}-20x+4-\left(2x-1\right)\left(2x+1\right)=47+x$$

Whakaarohia te \left(2x-1\right)\left(2x+1\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.

$$25x^{2}-20x+4-\left(\left(2x\right)^{2}-1\right)=47+x$$

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

$$25x^{2}-20x+4-\left(2^{2}x^{2}-1\right)=47+x$$

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

$$25x^{2}-20x+4-\left(4x^{2}-1\right)=47+x$$

Hei kimi i te tauaro o 4x^{2}-1, kimihia te tauaro o ia taurangi.

$$25x^{2}-20x+4-4x^{2}+1=47+x$$

Pahekotia te 25x^{2} me -4x^{2}, ka 21x^{2}.

$$21x^{2}-20x+4+1=47+x$$

Tāpirihia te 4 ki te 1, ka 5.

$$21x^{2}-20x+5=47+x$$

Tangohia te x mai i ngā taha e rua.

$$21x^{2}-20x+5-x=47$$

Pahekotia te -20x me -x, ka -21x.

$$21x^{2}-21x+5=47$$

Tangohia te 5 mai i ngā taha e rua.

$$21x^{2}-21x=47-5$$

Tangohia te 5 i te 47, ka 42.

$$21x^{2}-21x=42$$

Whakawehea ngā taha e rua ki te 21.

$$\frac{21x^{2}-21x}{21}=\frac{42}{21}$$

Mā te whakawehe ki te 21 ka wetekia te whakareanga ki te 21.

$$x^{2}+\left(-\frac{21}{21}\right)x=\frac{42}{21}$$

Whakawehe -21 ki te 21.

$$x^{2}-x=\frac{42}{21}$$

Whakawehe 42 ki te 21.

$$x^{2}-x=2$$

Whakawehea te -1, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{2}. Nā, tāpiria te pūrua o te -\frac{1}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-x+\frac{1}{4}=2+\frac{1}{4}$$

Tāpiri 2 ki te \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

Tauwehea x^{2}-x+\frac{1}{4}. 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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{3}{2} x-\frac{1}{2}=-\frac{3}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=2 x=-1$$