Réitigh (x-7)^2-x(6+x)operatorname{mon}x=-1/20

$$( x - 7 ) ^ { 2 } - x ( 6 + x ) \operatorname { mon } x = - \frac { 1 } { 20 }$$

Réitigh do m. (complex solution)

$$\left\{\begin{matrix}m=\frac{20x^{2}-280x+981}{20no\left(x+6\right)x^{2}}\text{, }&x\neq 0\text{ and }o\neq 0\text{ and }x\neq -6\text{ and }n\neq 0\\m\in \mathrm{C}\text{, }&\left(x=\frac{\sqrt{5}i}{10}+7\text{ and }n=0\right)\text{ or }\left(x=\frac{\sqrt{5}i}{10}+7\text{ and }o=0\right)\text{ or }\left(x=-\frac{\sqrt{5}i}{10}+7\text{ and }n=0\right)\text{ or }\left(x=-\frac{\sqrt{5}i}{10}+7\text{ and }o=0\right)\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh x agus x chun x^{2} a fháil.

$$\left(x-7\right)^{2}-x^{2}\left(6+x\right)mon=-\frac{1}{20}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(x-7\right)^{2} a leathnú.

$$x^{2}-14x+49-x^{2}\left(6+x\right)mon=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun x^{2} a mhéadú faoi 6+x.

$$x^{2}-14x+49-\left(6x^{2}+x^{3}\right)mon=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}+x^{3} a mhéadú faoi m.

$$x^{2}-14x+49-\left(6x^{2}m+x^{3}m\right)on=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}m+x^{3}m a mhéadú faoi o.

$$x^{2}-14x+49-\left(6x^{2}mo+x^{3}mo\right)n=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}mo+x^{3}mo a mhéadú faoi n.

$$x^{2}-14x+49-\left(6x^{2}mon+x^{3}mon\right)=-\frac{1}{20}$$

Chun an mhalairt ar 6x^{2}mon+x^{3}mon a aimsiú, aimsigh an mhalairt ar gach téarma.

$$x^{2}-14x+49-6x^{2}mon-x^{3}mon=-\frac{1}{20}$$

Bain x^{2} ón dá thaobh.

$$-14x+49-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}$$

Cuir 14x leis an dá thaobh.

$$49-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}+14x$$

Bain 49 ón dá thaobh.

$$-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}+14x-49$$

Dealaigh 49 ó -\frac{1}{20} chun -\frac{981}{20} a fháil.

$$-6x^{2}mon-x^{3}mon=-\frac{981}{20}-x^{2}+14x$$

Comhcheangail na téarmaí ar fad ina bhfuil m.

$$\left(-6x^{2}on-x^{3}on\right)m=-\frac{981}{20}-x^{2}+14x$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(-nox^{3}-6nox^{2}\right)m=-x^{2}+14x-\frac{981}{20}$$

Roinn an dá thaobh faoi -6x^{2}on-x^{3}on.

$$\frac{\left(-nox^{3}-6nox^{2}\right)m}{-nox^{3}-6nox^{2}}=\frac{-x^{2}+14x-\frac{981}{20}}{-nox^{3}-6nox^{2}}$$

Má roinntear é faoi -6x^{2}on-x^{3}on cuirtear an iolrúchán faoi -6x^{2}on-x^{3}on ar ceal.

$$m=\frac{-x^{2}+14x-\frac{981}{20}}{-nox^{3}-6nox^{2}}$$

Roinn -x^{2}+14x-\frac{981}{20} faoi -6x^{2}on-x^{3}on.

$$m=-\frac{-20x^{2}+280x-981}{20no\left(x+6\right)x^{2}}$$

Réitigh do n. (complex solution)

$$\left\{\begin{matrix}n=\frac{20x^{2}-280x+981}{20mo\left(x+6\right)x^{2}}\text{, }&x\neq 0\text{ and }o\neq 0\text{ and }x\neq -6\text{ and }m\neq 0\\n\in \mathrm{C}\text{, }&\left(x=\frac{\sqrt{5}i}{10}+7\text{ and }m=0\right)\text{ or }\left(x=\frac{\sqrt{5}i}{10}+7\text{ and }o=0\right)\text{ or }\left(x=-\frac{\sqrt{5}i}{10}+7\text{ and }m=0\right)\text{ or }\left(x=-\frac{\sqrt{5}i}{10}+7\text{ and }o=0\right)\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh x agus x chun x^{2} a fháil.

$$\left(x-7\right)^{2}-x^{2}\left(6+x\right)mon=-\frac{1}{20}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(x-7\right)^{2} a leathnú.

$$x^{2}-14x+49-x^{2}\left(6+x\right)mon=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun x^{2} a mhéadú faoi 6+x.

$$x^{2}-14x+49-\left(6x^{2}+x^{3}\right)mon=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}+x^{3} a mhéadú faoi m.

$$x^{2}-14x+49-\left(6x^{2}m+x^{3}m\right)on=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}m+x^{3}m a mhéadú faoi o.

$$x^{2}-14x+49-\left(6x^{2}mo+x^{3}mo\right)n=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}mo+x^{3}mo a mhéadú faoi n.

$$x^{2}-14x+49-\left(6x^{2}mon+x^{3}mon\right)=-\frac{1}{20}$$

Chun an mhalairt ar 6x^{2}mon+x^{3}mon a aimsiú, aimsigh an mhalairt ar gach téarma.

$$x^{2}-14x+49-6x^{2}mon-x^{3}mon=-\frac{1}{20}$$

Bain x^{2} ón dá thaobh.

$$-14x+49-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}$$

Cuir 14x leis an dá thaobh.

$$49-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}+14x$$

Bain 49 ón dá thaobh.

$$-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}+14x-49$$

Dealaigh 49 ó -\frac{1}{20} chun -\frac{981}{20} a fháil.

$$-6x^{2}mon-x^{3}mon=-\frac{981}{20}-x^{2}+14x$$

Comhcheangail na téarmaí ar fad ina bhfuil n.

$$\left(-6x^{2}mo-x^{3}mo\right)n=-\frac{981}{20}-x^{2}+14x$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(-mox^{3}-6mox^{2}\right)n=-x^{2}+14x-\frac{981}{20}$$

Roinn an dá thaobh faoi -6x^{2}mo-x^{3}mo.

$$\frac{\left(-mox^{3}-6mox^{2}\right)n}{-mox^{3}-6mox^{2}}=\frac{-x^{2}+14x-\frac{981}{20}}{-mox^{3}-6mox^{2}}$$

Má roinntear é faoi -6x^{2}mo-x^{3}mo cuirtear an iolrúchán faoi -6x^{2}mo-x^{3}mo ar ceal.

$$n=\frac{-x^{2}+14x-\frac{981}{20}}{-mox^{3}-6mox^{2}}$$

Roinn -x^{2}+14x-\frac{981}{20} faoi -6x^{2}mo-x^{3}mo.

$$n=-\frac{-20x^{2}+280x-981}{20mo\left(x+6\right)x^{2}}$$

Réitigh do m.

$$m=\frac{20x^{2}-280x+981}{20no\left(x+6\right)x^{2}}$$

$$x\neq 0\text{ and }o\neq 0\text{ and }x\neq -6\text{ and }n\neq 0$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh x agus x chun x^{2} a fháil.

$$\left(x-7\right)^{2}-x^{2}\left(6+x\right)mon=-\frac{1}{20}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(x-7\right)^{2} a leathnú.

$$x^{2}-14x+49-x^{2}\left(6+x\right)mon=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun x^{2} a mhéadú faoi 6+x.

$$x^{2}-14x+49-\left(6x^{2}+x^{3}\right)mon=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}+x^{3} a mhéadú faoi m.

$$x^{2}-14x+49-\left(6x^{2}m+x^{3}m\right)on=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}m+x^{3}m a mhéadú faoi o.

$$x^{2}-14x+49-\left(6x^{2}mo+x^{3}mo\right)n=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}mo+x^{3}mo a mhéadú faoi n.

$$x^{2}-14x+49-\left(6x^{2}mon+x^{3}mon\right)=-\frac{1}{20}$$

Chun an mhalairt ar 6x^{2}mon+x^{3}mon a aimsiú, aimsigh an mhalairt ar gach téarma.

$$x^{2}-14x+49-6x^{2}mon-x^{3}mon=-\frac{1}{20}$$

Bain x^{2} ón dá thaobh.

$$-14x+49-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}$$

Cuir 14x leis an dá thaobh.

$$49-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}+14x$$

Bain 49 ón dá thaobh.

$$-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}+14x-49$$

Dealaigh 49 ó -\frac{1}{20} chun -\frac{981}{20} a fháil.

$$-6x^{2}mon-x^{3}mon=-\frac{981}{20}-x^{2}+14x$$

Comhcheangail na téarmaí ar fad ina bhfuil m.

$$\left(-6x^{2}on-x^{3}on\right)m=-\frac{981}{20}-x^{2}+14x$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(-nox^{3}-6nox^{2}\right)m=-x^{2}+14x-\frac{981}{20}$$

Roinn an dá thaobh faoi -6x^{2}on-x^{3}on.

$$\frac{\left(-nox^{3}-6nox^{2}\right)m}{-nox^{3}-6nox^{2}}=\frac{-x^{2}+14x-\frac{981}{20}}{-nox^{3}-6nox^{2}}$$

Má roinntear é faoi -6x^{2}on-x^{3}on cuirtear an iolrúchán faoi -6x^{2}on-x^{3}on ar ceal.

$$m=\frac{-x^{2}+14x-\frac{981}{20}}{-nox^{3}-6nox^{2}}$$

Roinn -\frac{981}{20}-x^{2}+14x faoi -6x^{2}on-x^{3}on.

$$m=\frac{-20x^{2}+280x-981}{-20no\left(x+6\right)x^{2}}$$

Réitigh do n.

$$n=\frac{20x^{2}-280x+981}{20mo\left(x+6\right)x^{2}}$$

$$x\neq 0\text{ and }o\neq 0\text{ and }x\neq -6\text{ and }m\neq 0$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh x agus x chun x^{2} a fháil.

$$\left(x-7\right)^{2}-x^{2}\left(6+x\right)mon=-\frac{1}{20}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(x-7\right)^{2} a leathnú.

$$x^{2}-14x+49-x^{2}\left(6+x\right)mon=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun x^{2} a mhéadú faoi 6+x.

$$x^{2}-14x+49-\left(6x^{2}+x^{3}\right)mon=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}+x^{3} a mhéadú faoi m.

$$x^{2}-14x+49-\left(6x^{2}m+x^{3}m\right)on=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}m+x^{3}m a mhéadú faoi o.

$$x^{2}-14x+49-\left(6x^{2}mo+x^{3}mo\right)n=-\frac{1}{20}$$

Úsáid an t-airí dáileach chun 6x^{2}mo+x^{3}mo a mhéadú faoi n.

$$x^{2}-14x+49-\left(6x^{2}mon+x^{3}mon\right)=-\frac{1}{20}$$

Chun an mhalairt ar 6x^{2}mon+x^{3}mon a aimsiú, aimsigh an mhalairt ar gach téarma.

$$x^{2}-14x+49-6x^{2}mon-x^{3}mon=-\frac{1}{20}$$

Bain x^{2} ón dá thaobh.

$$-14x+49-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}$$

Cuir 14x leis an dá thaobh.

$$49-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}+14x$$

Bain 49 ón dá thaobh.

$$-6x^{2}mon-x^{3}mon=-\frac{1}{20}-x^{2}+14x-49$$

Dealaigh 49 ó -\frac{1}{20} chun -\frac{981}{20} a fháil.

$$-6x^{2}mon-x^{3}mon=-\frac{981}{20}-x^{2}+14x$$

Comhcheangail na téarmaí ar fad ina bhfuil n.

$$\left(-6x^{2}mo-x^{3}mo\right)n=-\frac{981}{20}-x^{2}+14x$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(-mox^{3}-6mox^{2}\right)n=-x^{2}+14x-\frac{981}{20}$$

Roinn an dá thaobh faoi -6x^{2}mo-x^{3}mo.

$$\frac{\left(-mox^{3}-6mox^{2}\right)n}{-mox^{3}-6mox^{2}}=\frac{-x^{2}+14x-\frac{981}{20}}{-mox^{3}-6mox^{2}}$$

Má roinntear é faoi -6x^{2}mo-x^{3}mo cuirtear an iolrúchán faoi -6x^{2}mo-x^{3}mo ar ceal.

$$n=\frac{-x^{2}+14x-\frac{981}{20}}{-mox^{3}-6mox^{2}}$$

Roinn -\frac{981}{20}-x^{2}+14x faoi -6x^{2}mo-x^{3}mo.

$$n=\frac{-20x^{2}+280x-981}{-20mo\left(x+6\right)x^{2}}$$