Réitigh 7x/x^2+4x+9+1/x^2-9=

$$\frac { 7 x } { x ^ { 2 } + 4 x + 9 } + \frac { 1 } { x ^ { 2 } - 9 } =$$

Luacháil

$$\frac{7x^{3}+x^{2}-59x+9}{\left(x^{2}-9\right)\left(x^{2}+4x+9\right)}$$

Céimeanna gearra réitigh

Fachtóirigh x^{2}-9.

$$\frac{7x}{x^{2}+4x+9}+\frac{1}{\left(x-3\right)\left(x+3\right)}$$

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de x^{2}+4x+9 agus \left(x-3\right)\left(x+3\right) ná \left(x-3\right)\left(x+3\right)\left(x^{2}+4x+9\right). Méadaigh \frac{7x}{x^{2}+4x+9} faoi \frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}. Méadaigh \frac{1}{\left(x-3\right)\left(x+3\right)} faoi \frac{x^{2}+4x+9}{x^{2}+4x+9}.

$$\frac{7x\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x^{2}+4x+9\right)}+\frac{x^{2}+4x+9}{\left(x-3\right)\left(x+3\right)\left(x^{2}+4x+9\right)}$$

Tá an t-ainmneoir céanna ag \frac{7x\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x^{2}+4x+9\right)} agus \frac{x^{2}+4x+9}{\left(x-3\right)\left(x+3\right)\left(x^{2}+4x+9\right)} agus, mar sin, is féidir iad a shuimiú trína n-uimhreoirí a shuimiú.

$$\frac{7x\left(x-3\right)\left(x+3\right)+x^{2}+4x+9}{\left(x-3\right)\left(x+3\right)\left(x^{2}+4x+9\right)}$$

Déan iolrúcháin in 7x\left(x-3\right)\left(x+3\right)+x^{2}+4x+9.

$$\frac{7x^{3}+21x^{2}-21x^{2}-63x+x^{2}+4x+9}{\left(x-3\right)\left(x+3\right)\left(x^{2}+4x+9\right)}$$

Cumaisc téarmaí comhchosúla in: 7x^{3}+21x^{2}-21x^{2}-63x+x^{2}+4x+9.

$$\frac{7x^{3}+x^{2}-59x+9}{\left(x-3\right)\left(x+3\right)\left(x^{2}+4x+9\right)}$$

Fairsingigh \left(x-3\right)\left(x+3\right)\left(x^{2}+4x+9\right)

$$\frac{7x^{3}+x^{2}-59x+9}{x^{4}+4x^{3}-36x-81}$$

Difreálaigh w.r.t. x

$$\frac{5103-162x-1845x^{2}-68x^{3}+173x^{4}-2x^{5}-7x^{6}}{\left(x^{2}-9\right)^{2}\left(x^{4}+8x^{3}+34x^{2}+72x+81\right)}$$