Réitigh (2x^2-1/(x-2)(x+1))^2-8(2x^2-1)+7.(x-1)(x+2)

$$( 2 x ^ { 2 } - \frac { 1 } { ( x - 2 ) ( x + 1 ) } ) ^ { 2 } - 8 ( 2 x ^ { 2 } - 1 ) + 7 . ( x - 1 ) ( x + 2 )$$

Luacháil

$$\frac{4x^{8}-8x^{7}-28x^{6}+48x^{5}+75x^{4}-90x^{3}-101x^{2}+60x+61}{\left(\left(x-2\right)\left(x+1\right)\right)^{2}}$$

Céimeanna gearra réitigh

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh 2x^{2} faoi \frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}.

$$\left(\frac{2x^{2}\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}-\frac{1}{\left(x-2\right)\left(x+1\right)}\right)^{2}-8\left(2x^{2}-1\right)+7$$

Tá an t-ainmneoir céanna ag \frac{2x^{2}\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)} agus \frac{1}{\left(x-2\right)\left(x+1\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

$$\left(\frac{2x^{2}\left(x-2\right)\left(x+1\right)-1}{\left(x-2\right)\left(x+1\right)}\right)^{2}-8\left(2x^{2}-1\right)+7$$

Déan iolrúcháin in 2x^{2}\left(x-2\right)\left(x+1\right)-1.

$$\left(\frac{2x^{4}+2x^{3}-4x^{3}-4x^{2}-1}{\left(x-2\right)\left(x+1\right)}\right)^{2}-8\left(2x^{2}-1\right)+7$$

Cumaisc téarmaí comhchosúla in: 2x^{4}+2x^{3}-4x^{3}-4x^{2}-1.

$$\left(\frac{2x^{4}-2x^{3}-4x^{2}-1}{\left(x-2\right)\left(x+1\right)}\right)^{2}-8\left(2x^{2}-1\right)+7$$

Chun \frac{2x^{4}-2x^{3}-4x^{2}-1}{\left(x-2\right)\left(x+1\right)} a iolrú i gcumhacht, iolraigh an t-uimhreoir agus an t-ainmneoir araon i gcumhacht agus déan iad a roinnt ansin.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(\left(x-2\right)\left(x+1\right)\right)^{2}}-8\left(2x^{2}-1\right)+7$$

Fairsingigh \left(\left(x-2\right)\left(x+1\right)\right)^{2}

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}-8\left(2x^{2}-1\right)+7$$

Úsáid an t-airí dáileach chun -8 a mhéadú faoi 2x^{2}-1.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}-16x^{2}+8+7$$

Suimigh 8 agus 7 chun 15 a fháil.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}-16x^{2}+15$$

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh -16x^{2}+15 faoi \frac{\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}+\frac{\left(-16x^{2}+15\right)\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}$$

Tá an t-ainmneoir céanna ag \frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}} agus \frac{\left(-16x^{2}+15\right)\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}} agus, mar sin, is féidir iad a shuimiú trína n-uimhreoirí a shuimiú.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}+\left(-16x^{2}+15\right)\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}$$

Déan iolrúcháin in \left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}+\left(-16x^{2}+15\right)\left(x-2\right)^{2}\left(x+1\right)^{2}.

$$\frac{4x^{8}-4x^{7}-8x^{6}-2x^{4}-4x^{7}+4x^{6}+8x^{5}+2x^{3}-8x^{6}+8x^{5}+16x^{4}+4x^{2}-2x^{4}+2x^{3}+4x^{2}+1-16x^{6}+32x^{5}+48x^{4}-64x^{3}-64x^{2}+15x^{4}-30x^{3}-45x^{2}+60x+60}{\left(x-2\right)^{2}\left(x+1\right)^{2}}$$

Cumaisc téarmaí comhchosúla in: 4x^{8}-4x^{7}-8x^{6}-2x^{4}-4x^{7}+4x^{6}+8x^{5}+2x^{3}-8x^{6}+8x^{5}+16x^{4}+4x^{2}-2x^{4}+2x^{3}+4x^{2}+1-16x^{6}+32x^{5}+48x^{4}-64x^{3}-64x^{2}+15x^{4}-30x^{3}-45x^{2}+60x+60.

$$\frac{4x^{8}-8x^{7}-28x^{6}+75x^{4}+48x^{5}-90x^{3}-101x^{2}+61+60x}{\left(x-2\right)^{2}\left(x+1\right)^{2}}$$

Fairsingigh \left(x-2\right)^{2}\left(x+1\right)^{2}

$$\frac{4x^{8}-8x^{7}-28x^{6}+75x^{4}+48x^{5}-90x^{3}-101x^{2}+61+60x}{x^{4}-2x^{3}-3x^{2}+4x+4}$$

Fairsingigh

$$\frac{4x^{8}-8x^{7}-28x^{6}+48x^{5}+75x^{4}-90x^{3}-101x^{2}+60x+61}{\left(\left(x-2\right)\left(x+1\right)\right)^{2}}$$

Céimeanna gearra réitigh

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh 2x^{2} faoi \frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}.

$$\left(\frac{2x^{2}\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}-\frac{1}{\left(x-2\right)\left(x+1\right)}\right)^{2}-8\left(2x^{2}-1\right)+7$$

Tá an t-ainmneoir céanna ag \frac{2x^{2}\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)} agus \frac{1}{\left(x-2\right)\left(x+1\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

$$\left(\frac{2x^{2}\left(x-2\right)\left(x+1\right)-1}{\left(x-2\right)\left(x+1\right)}\right)^{2}-8\left(2x^{2}-1\right)+7$$

Déan iolrúcháin in 2x^{2}\left(x-2\right)\left(x+1\right)-1.

$$\left(\frac{2x^{4}+2x^{3}-4x^{3}-4x^{2}-1}{\left(x-2\right)\left(x+1\right)}\right)^{2}-8\left(2x^{2}-1\right)+7$$

Cumaisc téarmaí comhchosúla in: 2x^{4}+2x^{3}-4x^{3}-4x^{2}-1.

$$\left(\frac{2x^{4}-2x^{3}-4x^{2}-1}{\left(x-2\right)\left(x+1\right)}\right)^{2}-8\left(2x^{2}-1\right)+7$$

Chun \frac{2x^{4}-2x^{3}-4x^{2}-1}{\left(x-2\right)\left(x+1\right)} a iolrú i gcumhacht, iolraigh an t-uimhreoir agus an t-ainmneoir araon i gcumhacht agus déan iad a roinnt ansin.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(\left(x-2\right)\left(x+1\right)\right)^{2}}-8\left(2x^{2}-1\right)+7$$

Fairsingigh \left(\left(x-2\right)\left(x+1\right)\right)^{2}

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}-8\left(2x^{2}-1\right)+7$$

Úsáid an t-airí dáileach chun -8 a mhéadú faoi 2x^{2}-1.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}-16x^{2}+8+7$$

Suimigh 8 agus 7 chun 15 a fháil.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}-16x^{2}+15$$

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh -16x^{2}+15 faoi \frac{\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}+\frac{\left(-16x^{2}+15\right)\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}$$

Tá an t-ainmneoir céanna ag \frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}} agus \frac{\left(-16x^{2}+15\right)\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}} agus, mar sin, is féidir iad a shuimiú trína n-uimhreoirí a shuimiú.

$$\frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}+\left(-16x^{2}+15\right)\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}$$

Déan iolrúcháin in \left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}+\left(-16x^{2}+15\right)\left(x-2\right)^{2}\left(x+1\right)^{2}.

$$\frac{4x^{8}-4x^{7}-8x^{6}-2x^{4}-4x^{7}+4x^{6}+8x^{5}+2x^{3}-8x^{6}+8x^{5}+16x^{4}+4x^{2}-2x^{4}+2x^{3}+4x^{2}+1-16x^{6}+32x^{5}+48x^{4}-64x^{3}-64x^{2}+15x^{4}-30x^{3}-45x^{2}+60x+60}{\left(x-2\right)^{2}\left(x+1\right)^{2}}$$

Cumaisc téarmaí comhchosúla in: 4x^{8}-4x^{7}-8x^{6}-2x^{4}-4x^{7}+4x^{6}+8x^{5}+2x^{3}-8x^{6}+8x^{5}+16x^{4}+4x^{2}-2x^{4}+2x^{3}+4x^{2}+1-16x^{6}+32x^{5}+48x^{4}-64x^{3}-64x^{2}+15x^{4}-30x^{3}-45x^{2}+60x+60.

$$\frac{4x^{8}-8x^{7}-28x^{6}+75x^{4}+48x^{5}-90x^{3}-101x^{2}+61+60x}{\left(x-2\right)^{2}\left(x+1\right)^{2}}$$

Fairsingigh \left(x-2\right)^{2}\left(x+1\right)^{2}

$$\frac{4x^{8}-8x^{7}-28x^{6}+75x^{4}+48x^{5}-90x^{3}-101x^{2}+61+60x}{x^{4}-2x^{3}-3x^{2}+4x+4}$$