Réitigh y=(x-3)^2x_{1}

$$y = ( x - 3 ) ^ { 2 } x _ { 1 }$$

Réitigh do x_1. (complex solution)

$$\left\{\begin{matrix}x_{1}=\frac{y}{\left(x-3\right)^{2}}\text{, }&x\neq 3\\x_{1}\in \mathrm{C}\text{, }&y=0\text{ and }x=3\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(x-3\right)^{2} a leathnú.

$$y=\left(x^{2}-6x+9\right)x_{1}$$

Úsáid an t-airí dáileach chun x^{2}-6x+9 a mhéadú faoi x_{1}.

$$y=x^{2}x_{1}-6xx_{1}+9x_{1}$$

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$x^{2}x_{1}-6xx_{1}+9x_{1}=y$$

Comhcheangail na téarmaí ar fad ina bhfuil x_{1}.

$$\left(x^{2}-6x+9\right)x_{1}=y$$

Roinn an dá thaobh faoi x^{2}-6x+9.

$$\frac{\left(x^{2}-6x+9\right)x_{1}}{x^{2}-6x+9}=\frac{y}{x^{2}-6x+9}$$

Má roinntear é faoi x^{2}-6x+9 cuirtear an iolrúchán faoi x^{2}-6x+9 ar ceal.

$$x_{1}=\frac{y}{x^{2}-6x+9}$$

Roinn y faoi x^{2}-6x+9.

$$x_{1}=\frac{y}{\left(x-3\right)^{2}}$$

Réitigh do x_1.

$$\left\{\begin{matrix}x_{1}=\frac{y}{\left(x-3\right)^{2}}\text{, }&x\neq 3\\x_{1}\in \mathrm{R}\text{, }&y=0\text{ and }x=3\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(x-3\right)^{2} a leathnú.

$$y=\left(x^{2}-6x+9\right)x_{1}$$

Úsáid an t-airí dáileach chun x^{2}-6x+9 a mhéadú faoi x_{1}.

$$y=x^{2}x_{1}-6xx_{1}+9x_{1}$$

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$x^{2}x_{1}-6xx_{1}+9x_{1}=y$$

Comhcheangail na téarmaí ar fad ina bhfuil x_{1}.

$$\left(x^{2}-6x+9\right)x_{1}=y$$

Roinn an dá thaobh faoi x^{2}-6x+9.

$$\frac{\left(x^{2}-6x+9\right)x_{1}}{x^{2}-6x+9}=\frac{y}{x^{2}-6x+9}$$

Má roinntear é faoi x^{2}-6x+9 cuirtear an iolrúchán faoi x^{2}-6x+9 ar ceal.

$$x_{1}=\frac{y}{x^{2}-6x+9}$$

Roinn y faoi x^{2}-6x+9.

$$x_{1}=\frac{y}{\left(x-3\right)^{2}}$$

Réitigh do x. (complex solution)

$$\left\{\begin{matrix}x=3+x_{1}^{-\frac{1}{2}}\sqrt{y}\text{; }x=3-x_{1}^{-\frac{1}{2}}\sqrt{y}\text{, }&x_{1}\neq 0\\x\in \mathrm{C}\text{, }&y=0\text{ and }x_{1}=0\end{matrix}\right.$$

Réitigh do x.

$$\left\{\begin{matrix}x=-\sqrt{\frac{y}{x_{1}}}+3\text{; }x=\sqrt{\frac{y}{x_{1}}}+3\text{, }&y\leq 0\text{ and }x_{1}<0\\x=-\sqrt{\frac{y}{x_{1}}}+3\text{; }x=\sqrt{\frac{y}{x_{1}}}+3\text{, }&y\geq 0\text{ and }x_{1}>0\\x\in \mathrm{R}\text{, }&y=0\text{ and }x_{1}=0\end{matrix}\right.$$