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

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

Calcular

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

Pasos de solución curta

Para sumar ou restar expresións, expándeas para facer que os seus denominadores sexan iguais. Multiplica 2x^{2} por \frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}.

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

Dado que \frac{2x^{2}\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)} e \frac{1}{\left(x-2\right)\left(x+1\right)} teñen o mesmo denominador, réstaos mediante a resta dos seus numeradores.

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

Fai as multiplicacións en 2x^{2}\left(x-2\right)\left(x+1\right)-1.

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

Combina como termos en 2x^{4}+2x^{3}-4x^{3}-4x^{2}-1.

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

Para elevar \frac{2x^{4}-2x^{3}-4x^{2}-1}{\left(x-2\right)\left(x+1\right)} a unha potencia, eleva o numerador e o denominador á potencia e despois divide.

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

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

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

Usa a propiedade distributiva para multiplicar -8 por 2x^{2}-1.

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

Suma 8 e 7 para obter 15.

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

Para sumar ou restar expresións, expándeas para facer que os seus denominadores sexan iguais. Multiplica -16x^{2}+15 por \frac{\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}.

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

Dado que \frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}} e \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}} teñen o mesmo denominador, súmaos mediante a suma dos seus numeradores.

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

Fai as multiplicacións en \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}.

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

Combina como termos en 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.

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

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

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

Expandir

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

Pasos de solución curta

Para sumar ou restar expresións, expándeas para facer que os seus denominadores sexan iguais. Multiplica 2x^{2} por \frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}.

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

Dado que \frac{2x^{2}\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)} e \frac{1}{\left(x-2\right)\left(x+1\right)} teñen o mesmo denominador, réstaos mediante a resta dos seus numeradores.

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

Fai as multiplicacións en 2x^{2}\left(x-2\right)\left(x+1\right)-1.

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

Combina como termos en 2x^{4}+2x^{3}-4x^{3}-4x^{2}-1.

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

Para elevar \frac{2x^{4}-2x^{3}-4x^{2}-1}{\left(x-2\right)\left(x+1\right)} a unha potencia, eleva o numerador e o denominador á potencia e despois divide.

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

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

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

Usa a propiedade distributiva para multiplicar -8 por 2x^{2}-1.

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

Suma 8 e 7 para obter 15.

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

Para sumar ou restar expresións, expándeas para facer que os seus denominadores sexan iguais. Multiplica -16x^{2}+15 por \frac{\left(x-2\right)^{2}\left(x+1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}}.

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

Dado que \frac{\left(2x^{4}-2x^{3}-4x^{2}-1\right)^{2}}{\left(x-2\right)^{2}\left(x+1\right)^{2}} e \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}} teñen o mesmo denominador, súmaos mediante a suma dos seus numeradores.

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

Fai as multiplicacións en \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}.

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

Combina como termos en 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.

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

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

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