Atrisiniet sqrt{m+1}+(n-3)^2=0

$$\sqrt { m + 1 } + ( n - 3 ) ^ { 2 } = 0$$

Atrast m

$$m=-1$$

$$n=3$$

Vienādojumu sakņu atrisināšanas darbības

Atņemiet \left(n-3\right)^{2} no vienādojuma abām pusēm.

$$\sqrt{m+1}+\left(n-3\right)^{2}-\left(n-3\right)^{2}=-\left(n-3\right)^{2}$$

Atņemot \left(n-3\right)^{2} no sevis, paliek 0.

$$\sqrt{m+1}=-\left(n-3\right)^{2}$$

Kāpiniet kvadrātā abas vienādojuma puses.

$$m+1=\left(n-3\right)^{4}$$

Atņemiet 1 no vienādojuma abām pusēm.

$$m+1-1=\left(n-3\right)^{4}-1$$

Atņemot 1 no sevis, paliek 0.

$$m=\left(n-3\right)^{4}-1$$

Atņemiet 1 no \left(n-3\right)^{4}.

$$m=\left(n-4\right)\left(n-2\right)\left(\left(n-3\right)^{2}+1\right)$$

Atrast m (complex solution)

$$m=-\left(4-n\right)\left(n-2\right)\left(\left(n-3\right)^{2}+1\right)$$

$$arg(\left(n-3\right)^{2})\geq \pi \text{ or }n=3$$

Atrast n (complex solution)

$$n=-i\sqrt[4]{m+1}+3$$

$$n=i\sqrt[4]{m+1}+3$$

Atrast n

$$n=3$$

$$m=-1$$