Question

Ayuda...
Necesito problemas resueltos de polimonios

pero para ya..

Answer 1

Respuesta:

Calcular:

1 P(x) + Q (x)

2 P(x) - U (x)  

3 P(x) + R (x)  

4 2P(x) - R (x)

5 S(x) + T(x) + U(x)

6 S(x) - T(x) + U(x)

Explicación paso a paso:

1 P(x) + Q (x)=

(4x^{2}-1) + (x^{3}-3x^{2} + 6x-2) =

= x^{3}-3x^{2} + 4x^{2} + 6x - 2 - 1 =

= x^{3} + x^{2} + 6x -3

 

2 P(x) - U (x) =

= (4x^{2} - 1) - (x^{2} + 2) =

= 4x^{2} - 1 - x^{2} - 2 =

= 3x^{2} - 3

 

3 P(x) + R (x) =

= (4x^{2} - 1) + (6x^{2} + x + 1) =

= 4x^{2} + 6x^{2} + x - 1 + 1 =

= 10x^{2} + x

 

4 2P(x) - R (x) =

= 2 \cdot (4x^{2} - 1) - (6x^{2} + x + 1) =

= 8x^{2} - 2 - 6x^{2} - x - 1 =

=2x^{2} - x - 3

 

5 S(x) + T(x) + U(x) =

= \displaystyle (\frac{1}{2x^{2}} + 4) + (\frac{3}{2x^{2}} + 5) + (x^{2} + 2) =

= \displaystyle \frac{1}{2x^{2}} + \frac{3}{2x^{2}} + x^{2} + 4 + 5 + 2 =

= x^{2}+\frac{2}{x^{2}}+11

 

6 S(x) - T(x) + U(x) =

= \displaystyle (\frac{1}{2x^{2}} + 4) - (\frac{3}{2x^{2}} + 5) + (x^{2} + 2) =

= \displaystyle \frac{1}{2x^{2}} + 4 - \frac{3}{2x^{2}} -  5 + x^{2} + 2 =

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