Réitigh 4x-3/2x+1-10(2x-1/4x-3)

$$\frac { 4 x - 3 } { 2 x + 1 } - 10 ( \frac { 2 x - 1 } { 4 x - 3 } )$$

Luacháil

$$-\frac{24x^{2}+24x-19}{\left(4x-3\right)\left(2x+1\right)}$$

Céimeanna gearra réitigh

Scríobh 10\times \frac{2x-1}{4x-3} mar chodán aonair.

$$\frac{4x-3}{2x+1}-\frac{10\left(2x-1\right)}{4x-3}$$

Úsáid an t-airí dáileach chun 10 a mhéadú faoi 2x-1.

$$\frac{4x-3}{2x+1}-\frac{20x-10}{4x-3}$$

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de 2x+1 agus 4x-3 ná \left(4x-3\right)\left(2x+1\right). Méadaigh \frac{4x-3}{2x+1} faoi \frac{4x-3}{4x-3}. Méadaigh \frac{20x-10}{4x-3} faoi \frac{2x+1}{2x+1}.

$$\frac{\left(4x-3\right)\left(4x-3\right)}{\left(4x-3\right)\left(2x+1\right)}-\frac{\left(20x-10\right)\left(2x+1\right)}{\left(4x-3\right)\left(2x+1\right)}$$

Tá an t-ainmneoir céanna ag \frac{\left(4x-3\right)\left(4x-3\right)}{\left(4x-3\right)\left(2x+1\right)} agus \frac{\left(20x-10\right)\left(2x+1\right)}{\left(4x-3\right)\left(2x+1\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

$$\frac{\left(4x-3\right)\left(4x-3\right)-\left(20x-10\right)\left(2x+1\right)}{\left(4x-3\right)\left(2x+1\right)}$$

Déan iolrúcháin in \left(4x-3\right)\left(4x-3\right)-\left(20x-10\right)\left(2x+1\right).

$$\frac{16x^{2}-12x-12x+9-40x^{2}-20x+20x+10}{\left(4x-3\right)\left(2x+1\right)}$$

Cumaisc téarmaí comhchosúla in: 16x^{2}-12x-12x+9-40x^{2}-20x+20x+10.

$$\frac{-24x^{2}-24x+19}{\left(4x-3\right)\left(2x+1\right)}$$

Fairsingigh \left(4x-3\right)\left(2x+1\right)

$$\frac{-24x^{2}-24x+19}{8x^{2}-2x-3}$$

Fairsingigh

$$\frac{19-24x-24x^{2}}{\left(4x-3\right)\left(2x+1\right)}$$

Céimeanna gearra réitigh

Scríobh 10\times \frac{2x-1}{4x-3} mar chodán aonair.

$$\frac{4x-3}{2x+1}-\frac{10\left(2x-1\right)}{4x-3}$$

Úsáid an t-airí dáileach chun 10 a mhéadú faoi 2x-1.

$$\frac{4x-3}{2x+1}-\frac{20x-10}{4x-3}$$

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de 2x+1 agus 4x-3 ná \left(4x-3\right)\left(2x+1\right). Méadaigh \frac{4x-3}{2x+1} faoi \frac{4x-3}{4x-3}. Méadaigh \frac{20x-10}{4x-3} faoi \frac{2x+1}{2x+1}.

$$\frac{\left(4x-3\right)\left(4x-3\right)}{\left(4x-3\right)\left(2x+1\right)}-\frac{\left(20x-10\right)\left(2x+1\right)}{\left(4x-3\right)\left(2x+1\right)}$$

Tá an t-ainmneoir céanna ag \frac{\left(4x-3\right)\left(4x-3\right)}{\left(4x-3\right)\left(2x+1\right)} agus \frac{\left(20x-10\right)\left(2x+1\right)}{\left(4x-3\right)\left(2x+1\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

$$\frac{\left(4x-3\right)\left(4x-3\right)-\left(20x-10\right)\left(2x+1\right)}{\left(4x-3\right)\left(2x+1\right)}$$

Déan iolrúcháin in \left(4x-3\right)\left(4x-3\right)-\left(20x-10\right)\left(2x+1\right).

$$\frac{16x^{2}-12x-12x+9-40x^{2}-20x+20x+10}{\left(4x-3\right)\left(2x+1\right)}$$

Cumaisc téarmaí comhchosúla in: 16x^{2}-12x-12x+9-40x^{2}-20x+20x+10.

$$\frac{-24x^{2}-24x+19}{\left(4x-3\right)\left(2x+1\right)}$$

Fairsingigh \left(4x-3\right)\left(2x+1\right)

$$\frac{-24x^{2}-24x+19}{8x^{2}-2x-3}$$