Whakaoti i te 4(x^2+2x)+1/x^2+2x=-5

$$4 ( x ^ { 2 } + 2 x ) + \frac { 1 } { x ^ { 2 } + 2 x } = - 5$$

Whakaoti mō x

$$x=-\frac{\sqrt{3}}{2}-1\approx -1.866025404$$

$$x=\frac{\sqrt{3}}{2}-1\approx -0.133974596$$

$$x=-1$$

Ngā Upane Otinga

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x\left(x+2\right).

$$4\left(x^{2}+2x\right)x\left(x+2\right)+1=-5x\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{2}+2x.

$$\left(4x^{2}+8x\right)x\left(x+2\right)+1=-5x\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4x^{2}+8x ki te x.

$$\left(4x^{3}+8x^{2}\right)\left(x+2\right)+1=-5x\left(x+2\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x^{3}+8x^{2} ki te x+2 ka whakakotahi i ngā kupu rite.

$$4x^{4}+16x^{3}+16x^{2}+1=-5x\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -5x ki te x+2.

$$4x^{4}+16x^{3}+16x^{2}+1=-5x^{2}-10x$$

Me tāpiri te 5x^{2} ki ngā taha e rua.

$$4x^{4}+16x^{3}+16x^{2}+1+5x^{2}=-10x$$

Pahekotia te 16x^{2} me 5x^{2}, ka 21x^{2}.

$$4x^{4}+16x^{3}+21x^{2}+1=-10x$$

Me tāpiri te 10x ki ngā taha e rua.

$$4x^{4}+16x^{3}+21x^{2}+1+10x=0$$

Hurinahatia te whārite ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$4x^{4}+16x^{3}+21x^{2}+10x+1=0$$

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau 1, ā, ka wehea e q te whakarea arahanga 4. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{1}{4},±\frac{1}{2},±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$x=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 4x^{4}+16x^{3}+21x^{2}+10x+1 ki te x+1, kia riro ko 4x^{3}+12x^{2}+9x+1. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$4x^{3}+12x^{2}+9x+1=0$$

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau 1, ā, ka wehea e q te whakarea arahanga 4. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{1}{4},±\frac{1}{2},±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$x=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 4x^{3}+12x^{2}+9x+1 ki te x+1, kia riro ko 4x^{2}+8x+1. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$4x^{2}+8x+1=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 4 mō te a, te 8 mō te b, me te 1 mō te c i te ture pūrua.

$$x=\frac{-8±\sqrt{8^{2}-4\times 4\times 1}}{2\times 4}$$

Mahia ngā tātaitai.

$$x=\frac{-8±4\sqrt{3}}{8}$$

Whakaotia te whārite 4x^{2}+8x+1=0 ina he tōrunga te ±, ina he tōraro te ±.

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

Rārangitia ngā otinga katoa i kitea.

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