Réitigh (2+1/x)^2/1+x-(1-1/x)^2*(x+1/x-2)-frac{2x+1}{x^2+x}

$$\frac { ( 2 + \frac { 1 } { x } ) ^ { 2 } } { 1 + x } - ( 1 - \frac { 1 } { x } ) ^ { 2 } \cdot ( x + \frac { 1 } { x } - 2 ) - \frac { 2 x + 1 } { x ^ { 2 } + x }$$

Luacháil

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

Céimeanna gearra réitigh

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh 2 faoi \frac{x}{x}.

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

Tá an t-ainmneoir céanna ag \frac{2x}{x} agus \frac{1}{x} agus, mar sin, is féidir iad a shuimiú trína n-uimhreoirí a shuimiú.

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

Chun \frac{2x+1}{x} a iolrú i gcumhacht, iolraigh an t-uimhreoir agus an t-ainmneoir araon i gcumhacht agus déan iad a roinnt ansin.

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

Scríobh \frac{\frac{\left(2x+1\right)^{2}}{x^{2}}}{1+x} mar chodán aonair.

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

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh 1 faoi \frac{x}{x}.

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

Tá an t-ainmneoir céanna ag \frac{x}{x} agus \frac{1}{x} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

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

Chun \frac{x-1}{x} a iolrú i gcumhacht, iolraigh an t-uimhreoir agus an t-ainmneoir araon i gcumhacht agus déan iad a roinnt ansin.

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

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh x-2 faoi \frac{x}{x}.

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

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

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

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

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

Méadaigh \frac{\left(x-1\right)^{2}}{x^{2}} faoi \frac{x^{2}-2x+1}{x} tríd an uimhreoir a mhéadú faoin uimhreoir agus an t-ainmneoir a mhéadú faoin ainmneoir.

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

Chun cumhachtaí den bhonn céanna a iolrú, suimigh a n-easpónaint. Suimigh 2 agus 1 chun 3 a bhaint amach.

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

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}\left(1+x\right) agus x^{3} ná \left(x+1\right)x^{3}. Méadaigh \frac{\left(2x+1\right)^{2}}{x^{2}\left(1+x\right)} faoi \frac{x}{x}. Méadaigh \frac{\left(x-1\right)^{2}\left(x^{2}-2x+1\right)}{x^{3}} faoi \frac{x+1}{x+1}.

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

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

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

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

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

Cumaisc téarmaí comhchosúla in: 4x^{3}+4x^{2}+x-x^{5}+x^{4}+x^{3}-x^{2}+2x^{4}-2x^{3}-2x^{2}+2x-x^{3}+x^{2}+x-1.

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

Fachtóirigh x^{2}+x.

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

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

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

Tá an t-ainmneoir céanna ag \frac{2x^{3}+2x^{2}+4x-x^{5}+3x^{4}-1}{\left(x+1\right)x^{3}} agus \frac{\left(2x+1\right)x^{2}}{\left(x+1\right)x^{3}} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

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

Déan iolrúcháin in 2x^{3}+2x^{2}+4x-x^{5}+3x^{4}-1-\left(2x+1\right)x^{2}.

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

Cumaisc téarmaí comhchosúla in: 2x^{3}+2x^{2}+4x-x^{5}+3x^{4}-1-2x^{3}-x^{2}.

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

Fachtóirigh na sloinn nach bhfuil fachtóirithe cheana in \frac{x^{2}+4x-x^{5}+3x^{4}-1}{\left(x+1\right)x^{3}}.

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

Cealaigh x+1 mar uimhreoir agus ainmneoir.

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

Fairsingigh

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

Céimeanna gearra réitigh

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh 2 faoi \frac{x}{x}.

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

Tá an t-ainmneoir céanna ag \frac{2x}{x} agus \frac{1}{x} agus, mar sin, is féidir iad a shuimiú trína n-uimhreoirí a shuimiú.

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

Chun \frac{2x+1}{x} a iolrú i gcumhacht, iolraigh an t-uimhreoir agus an t-ainmneoir araon i gcumhacht agus déan iad a roinnt ansin.

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

Scríobh \frac{\frac{\left(2x+1\right)^{2}}{x^{2}}}{1+x} mar chodán aonair.

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

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh 1 faoi \frac{x}{x}.

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

Tá an t-ainmneoir céanna ag \frac{x}{x} agus \frac{1}{x} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

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

Chun \frac{x-1}{x} a iolrú i gcumhacht, iolraigh an t-uimhreoir agus an t-ainmneoir araon i gcumhacht agus déan iad a roinnt ansin.

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

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh x-2 faoi \frac{x}{x}.

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

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

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

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

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

Méadaigh \frac{\left(x-1\right)^{2}}{x^{2}} faoi \frac{x^{2}-2x+1}{x} tríd an uimhreoir a mhéadú faoin uimhreoir agus an t-ainmneoir a mhéadú faoin ainmneoir.

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

Chun cumhachtaí den bhonn céanna a iolrú, suimigh a n-easpónaint. Suimigh 2 agus 1 chun 3 a bhaint amach.

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

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}\left(1+x\right) agus x^{3} ná \left(x+1\right)x^{3}. Méadaigh \frac{\left(2x+1\right)^{2}}{x^{2}\left(1+x\right)} faoi \frac{x}{x}. Méadaigh \frac{\left(x-1\right)^{2}\left(x^{2}-2x+1\right)}{x^{3}} faoi \frac{x+1}{x+1}.

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

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

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

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

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

Cumaisc téarmaí comhchosúla in: 4x^{3}+4x^{2}+x-x^{5}+x^{4}+x^{3}-x^{2}+2x^{4}-2x^{3}-2x^{2}+2x-x^{3}+x^{2}+x-1.

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

Fachtóirigh x^{2}+x.

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

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

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

Tá an t-ainmneoir céanna ag \frac{2x^{3}+2x^{2}+4x-x^{5}+3x^{4}-1}{\left(x+1\right)x^{3}} agus \frac{\left(2x+1\right)x^{2}}{\left(x+1\right)x^{3}} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

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

Déan iolrúcháin in 2x^{3}+2x^{2}+4x-x^{5}+3x^{4}-1-\left(2x+1\right)x^{2}.

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

Cumaisc téarmaí comhchosúla in: 2x^{3}+2x^{2}+4x-x^{5}+3x^{4}-1-2x^{3}-x^{2}.

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

Fachtóirigh na sloinn nach bhfuil fachtóirithe cheana in \frac{x^{2}+4x-x^{5}+3x^{4}-1}{\left(x+1\right)x^{3}}.

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

Cealaigh x+1 mar uimhreoir agus ainmneoir.

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