Réitigh 3x^{2}+5x+7+2x^3+3x+16/x^2+x=10x^3+12x+4/x-2+7x^3/x+1

$$3 x ^ { 2 } + 5 x + 7 + \frac { 2 x ^ { 3 } + 3 x + 16 } { x ^ { 2 } + x } = \frac { 10 x ^ { 3 } + 12 x + 4 } { x } - \frac { 2 + 7 x ^ { 3 } } { x + 1 }$$

Réitigh do x.

$$x=3$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Ní féidir leis an athróg x a bheith comhionann le haon cheann de na luachanna -1,0 toisc nach bhfuil an roinnt faoi nialas sainithe. Iolraigh an dá thaobh den chothromóid faoi x\left(x+1\right), an comhiolraí is lú de x^{2}+x,x,x+1.

$$3x^{2}x\left(x+1\right)+5xx\left(x+1\right)+x\left(x+1\right)\times 7+2x^{3}+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

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

$$3x^{3}\left(x+1\right)+5xx\left(x+1\right)+x\left(x+1\right)\times 7+2x^{3}+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

Úsáid an t-airí dáileach chun 3x^{3} a mhéadú faoi x+1.

$$3x^{4}+3x^{3}+5xx\left(x+1\right)+x\left(x+1\right)\times 7+2x^{3}+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

Méadaigh x agus x chun x^{2} a fháil.

$$3x^{4}+3x^{3}+5x^{2}\left(x+1\right)+x\left(x+1\right)\times 7+2x^{3}+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

Úsáid an t-airí dáileach chun 5x^{2} a mhéadú faoi x+1.

$$3x^{4}+3x^{3}+5x^{3}+5x^{2}+x\left(x+1\right)\times 7+2x^{3}+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

Comhcheangail 3x^{3} agus 5x^{3} chun 8x^{3} a fháil.

$$3x^{4}+8x^{3}+5x^{2}+x\left(x+1\right)\times 7+2x^{3}+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

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

$$3x^{4}+8x^{3}+5x^{2}+\left(x^{2}+x\right)\times 7+2x^{3}+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

Úsáid an t-airí dáileach chun x^{2}+x a mhéadú faoi 7.

$$3x^{4}+8x^{3}+5x^{2}+7x^{2}+7x+2x^{3}+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

Comhcheangail 5x^{2} agus 7x^{2} chun 12x^{2} a fháil.

$$3x^{4}+8x^{3}+12x^{2}+7x+2x^{3}+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

Comhcheangail 8x^{3} agus 2x^{3} chun 10x^{3} a fháil.

$$3x^{4}+10x^{3}+12x^{2}+7x+3x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

Comhcheangail 7x agus 3x chun 10x a fháil.

$$3x^{4}+10x^{3}+12x^{2}+10x+16=\left(x+1\right)\left(10x^{3}+12x+4\right)-x\left(2+7x^{3}\right)$$

Úsáid an t-airí dáileach chun x+1 a mhéadú faoi 10x^{3}+12x+4 agus chun téarmaí comhchosúla a chumasc.

$$3x^{4}+10x^{3}+12x^{2}+10x+16=10x^{4}+12x^{2}+16x+10x^{3}+4-x\left(2+7x^{3}\right)$$

Úsáid an t-airí dáileach chun x a mhéadú faoi 2+7x^{3}.

$$3x^{4}+10x^{3}+12x^{2}+10x+16=10x^{4}+12x^{2}+16x+10x^{3}+4-\left(2x+7x^{4}\right)$$

Chun an mhalairt ar 2x+7x^{4} a aimsiú, aimsigh an mhalairt ar gach téarma.

$$3x^{4}+10x^{3}+12x^{2}+10x+16=10x^{4}+12x^{2}+16x+10x^{3}+4-2x-7x^{4}$$

Comhcheangail 16x agus -2x chun 14x a fháil.

$$3x^{4}+10x^{3}+12x^{2}+10x+16=10x^{4}+12x^{2}+14x+10x^{3}+4-7x^{4}$$

Comhcheangail 10x^{4} agus -7x^{4} chun 3x^{4} a fháil.

$$3x^{4}+10x^{3}+12x^{2}+10x+16=3x^{4}+12x^{2}+14x+10x^{3}+4$$

Bain 3x^{4} ón dá thaobh.

$$3x^{4}+10x^{3}+12x^{2}+10x+16-3x^{4}=12x^{2}+14x+10x^{3}+4$$

Comhcheangail 3x^{4} agus -3x^{4} chun 0 a fháil.

$$10x^{3}+12x^{2}+10x+16=12x^{2}+14x+10x^{3}+4$$

Bain 12x^{2} ón dá thaobh.

$$10x^{3}+12x^{2}+10x+16-12x^{2}=14x+10x^{3}+4$$

Comhcheangail 12x^{2} agus -12x^{2} chun 0 a fháil.

$$10x^{3}+10x+16=14x+10x^{3}+4$$

Bain 14x ón dá thaobh.

$$10x^{3}+10x+16-14x=10x^{3}+4$$

Comhcheangail 10x agus -14x chun -4x a fháil.

$$10x^{3}-4x+16=10x^{3}+4$$

Bain 10x^{3} ón dá thaobh.

$$10x^{3}-4x+16-10x^{3}=4$$

Comhcheangail 10x^{3} agus -10x^{3} chun 0 a fháil.

$$-4x+16=4$$

Bain 16 ón dá thaobh.

$$-4x=4-16$$

Dealaigh 16 ó 4 chun -12 a fháil.

$$-4x=-12$$

Roinn an dá thaobh faoi -4.

$$x=\frac{-12}{-4}$$

Roinn -12 faoi -4 chun 3 a fháil.

$$x=3$$