Lahenda dy/dydydx=1-1/sqrt{2}

$$\frac { d y } { d y } d y d x = 1 - \frac { 1 } { \sqrt { 2 } }$$

Lahendage ja leidke x

$$x=\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{2yd^{2}}$$

$$d\neq 0\text{ and }y\neq 0$$

Juhised lineaarvõrrandi lahendamiseks

Korrutage d ja d, et leida d^{2}.

$$\frac{\mathrm{d}(y)}{\mathrm{d}y}d^{2}yx=1-\frac{1}{\sqrt{2}}$$

Ratsionaliseerige korrutades lugeja ja \sqrt{2} nimetaja \frac{1}{\sqrt{2}} nimetaja.

$$\frac{\mathrm{d}(y)}{\mathrm{d}y}d^{2}yx=1-\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}$$

\sqrt{2} ruut on 2.

$$\frac{\mathrm{d}(y)}{\mathrm{d}y}d^{2}yx=1-\frac{\sqrt{2}}{2}$$

Korrutage võrrandi mõlemad pooled 2-ga.

$$2\frac{\mathrm{d}(y)}{\mathrm{d}y}d^{2}yx=2-\sqrt{2}$$

Võrrand on standardkujul.

$$2yd^{2}x=2-\sqrt{2}$$

Jagage mõlemad pooled 2d^{2}y-ga.

$$\frac{2yd^{2}x}{2yd^{2}}=\frac{2-\sqrt{2}}{2yd^{2}}$$

2d^{2}y-ga jagamine võtab 2d^{2}y-ga korrutamise tagasi.

$$x=\frac{2-\sqrt{2}}{2yd^{2}}$$

Lahendage ja leidke d (complex solution)

$$d=-\frac{\sqrt{4-2\sqrt{2}}x^{-\frac{1}{2}}y^{-\frac{1}{2}}}{2}$$

$$d=\frac{\sqrt{4-2\sqrt{2}}x^{-\frac{1}{2}}y^{-\frac{1}{2}}}{2}\text{, }x\neq 0\text{ and }y\neq 0$$

Lahendage ja leidke d

$$d=\frac{\sqrt{\frac{4-2\sqrt{2}}{xy}}}{2}$$

$$d=-\frac{\sqrt{\frac{4-2\sqrt{2}}{xy}}}{2}\text{, }\left(x>0\text{ and }y>0\right)\text{ or }\left(y<0\text{ and }x<0\right)$$