Lahenda (x-2y/x+2y+frac{x+2y/x-2y)*(1+frac{x^{2}+4y^2}{4xy})}{x^2+4y^2/2xy*(x^2+2xy)}

$$\frac { ( \frac { x - 2 y } { x + 2 y } + \frac { x + 2 y } { x - 2 y } ) \cdot ( 1 + \frac { x ^ { 2 } + 4 y ^ { 2 } } { 4 x y } ) } { \frac { x ^ { 2 } + 4 y ^ { 2 } } { 2 x y } \cdot ( x ^ { 2 } + 2 x y ) }$$

Arvuta

$$\frac{1}{x\left(x-2y\right)}$$

Lühike lahenduskäik

Avaldiste liitmiseks või lahutamiseks laiendage need, et neil oleksid ühised nimetajad. x+2y ja x-2y vähim ühiskordne on \left(x-2y\right)\left(x+2y\right). Korrutage omavahel \frac{x-2y}{x+2y} ja \frac{x-2y}{x-2y}. Korrutage omavahel \frac{x+2y}{x-2y} ja \frac{x+2y}{x+2y}.

$$\frac{\left(\frac{\left(x-2y\right)\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{\left(x+2y\right)\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}\right)\left(1+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Kuna murdudel \frac{\left(x-2y\right)\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)} ja \frac{\left(x+2y\right)\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)} on sama nimetaja, liitke nende lugejad.

$$\frac{\frac{\left(x-2y\right)\left(x-2y\right)+\left(x+2y\right)\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}\left(1+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Tehke korrutustehted võrrandis \left(x-2y\right)\left(x-2y\right)+\left(x+2y\right)\left(x+2y\right).

$$\frac{\frac{x^{2}-2xy-2xy+4y^{2}+x^{2}+2xy+2xy+4y^{2}}{\left(x-2y\right)\left(x+2y\right)}\left(1+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Kombineerige sarnased liikmed avaldises x^{2}-2xy-2xy+4y^{2}+x^{2}+2xy+2xy+4y^{2}.

$$\frac{\frac{2x^{2}+8y^{2}}{\left(x-2y\right)\left(x+2y\right)}\left(1+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Avaldiste liitmiseks või lahutamiseks laiendage need, et neil oleksid ühised nimetajad. Korrutage omavahel 1 ja \frac{4xy}{4xy}.

$$\frac{\frac{2x^{2}+8y^{2}}{\left(x-2y\right)\left(x+2y\right)}\left(\frac{4xy}{4xy}+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Kuna murdudel \frac{4xy}{4xy} ja \frac{x^{2}+4y^{2}}{4xy} on sama nimetaja, liitke nende lugejad.

$$\frac{\frac{2x^{2}+8y^{2}}{\left(x-2y\right)\left(x+2y\right)}\times \frac{4xy+x^{2}+4y^{2}}{4xy}}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Korrutage omavahel \frac{2x^{2}+8y^{2}}{\left(x-2y\right)\left(x+2y\right)} ja \frac{4xy+x^{2}+4y^{2}}{4xy}. Seejärel taandage murd võimaluse korral vähimale ühiskordsele.

$$\frac{\frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)}{\left(x-2y\right)\left(x+2y\right)\times 4xy}}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Avaldage \frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right) ühe murdarvuna.

$$\frac{\frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)}{\left(x-2y\right)\left(x+2y\right)\times 4xy}}{\frac{\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}{2xy}}$$

Jagage \frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)}{\left(x-2y\right)\left(x+2y\right)\times 4xy} väärtusega \frac{\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}{2xy}, korrutades \frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)}{\left(x-2y\right)\left(x+2y\right)\times 4xy} väärtuse \frac{\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}{2xy} pöördväärtusega.

$$\frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)\times 2xy}{\left(x-2y\right)\left(x+2y\right)\times 4xy\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}$$

Taandage 2xy nii lugejas kui ka nimetajas.

$$\frac{\left(2x^{2}+8y^{2}\right)\left(x^{2}+4xy+4y^{2}\right)}{2\left(x-2y\right)\left(x+2y\right)\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}$$

Kui avaldised pole juba teguriteks lahutatud, tehke seda.

$$\frac{2\left(x+2y\right)^{2}\left(x^{2}+4y^{2}\right)}{2x\left(x-2y\right)\left(x+2y\right)^{2}\left(x^{2}+4y^{2}\right)}$$

Taandage 2\left(x+2y\right)^{2}\left(x^{2}+4y^{2}\right) nii lugejas kui ka nimetajas.

$$\frac{1}{x\left(x-2y\right)}$$

Laiendage avaldist.

$$\frac{1}{x^{2}-2xy}$$

Laienda

$$\frac{1}{x\left(x-2y\right)}$$

Lühike lahenduskäik

Avaldiste liitmiseks või lahutamiseks laiendage need, et neil oleksid ühised nimetajad. x+2y ja x-2y vähim ühiskordne on \left(x-2y\right)\left(x+2y\right). Korrutage omavahel \frac{x-2y}{x+2y} ja \frac{x-2y}{x-2y}. Korrutage omavahel \frac{x+2y}{x-2y} ja \frac{x+2y}{x+2y}.

$$\frac{\left(\frac{\left(x-2y\right)\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{\left(x+2y\right)\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}\right)\left(1+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Kuna murdudel \frac{\left(x-2y\right)\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)} ja \frac{\left(x+2y\right)\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)} on sama nimetaja, liitke nende lugejad.

$$\frac{\frac{\left(x-2y\right)\left(x-2y\right)+\left(x+2y\right)\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}\left(1+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Tehke korrutustehted võrrandis \left(x-2y\right)\left(x-2y\right)+\left(x+2y\right)\left(x+2y\right).

$$\frac{\frac{x^{2}-2xy-2xy+4y^{2}+x^{2}+2xy+2xy+4y^{2}}{\left(x-2y\right)\left(x+2y\right)}\left(1+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Kombineerige sarnased liikmed avaldises x^{2}-2xy-2xy+4y^{2}+x^{2}+2xy+2xy+4y^{2}.

$$\frac{\frac{2x^{2}+8y^{2}}{\left(x-2y\right)\left(x+2y\right)}\left(1+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Avaldiste liitmiseks või lahutamiseks laiendage need, et neil oleksid ühised nimetajad. Korrutage omavahel 1 ja \frac{4xy}{4xy}.

$$\frac{\frac{2x^{2}+8y^{2}}{\left(x-2y\right)\left(x+2y\right)}\left(\frac{4xy}{4xy}+\frac{x^{2}+4y^{2}}{4xy}\right)}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Kuna murdudel \frac{4xy}{4xy} ja \frac{x^{2}+4y^{2}}{4xy} on sama nimetaja, liitke nende lugejad.

$$\frac{\frac{2x^{2}+8y^{2}}{\left(x-2y\right)\left(x+2y\right)}\times \frac{4xy+x^{2}+4y^{2}}{4xy}}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Korrutage omavahel \frac{2x^{2}+8y^{2}}{\left(x-2y\right)\left(x+2y\right)} ja \frac{4xy+x^{2}+4y^{2}}{4xy}. Seejärel taandage murd võimaluse korral vähimale ühiskordsele.

$$\frac{\frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)}{\left(x-2y\right)\left(x+2y\right)\times 4xy}}{\frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right)}$$

Avaldage \frac{x^{2}+4y^{2}}{2xy}\left(x^{2}+2xy\right) ühe murdarvuna.

$$\frac{\frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)}{\left(x-2y\right)\left(x+2y\right)\times 4xy}}{\frac{\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}{2xy}}$$

Jagage \frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)}{\left(x-2y\right)\left(x+2y\right)\times 4xy} väärtusega \frac{\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}{2xy}, korrutades \frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)}{\left(x-2y\right)\left(x+2y\right)\times 4xy} väärtuse \frac{\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}{2xy} pöördväärtusega.

$$\frac{\left(2x^{2}+8y^{2}\right)\left(4xy+x^{2}+4y^{2}\right)\times 2xy}{\left(x-2y\right)\left(x+2y\right)\times 4xy\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}$$

Taandage 2xy nii lugejas kui ka nimetajas.

$$\frac{\left(2x^{2}+8y^{2}\right)\left(x^{2}+4xy+4y^{2}\right)}{2\left(x-2y\right)\left(x+2y\right)\left(x^{2}+4y^{2}\right)\left(x^{2}+2xy\right)}$$

Kui avaldised pole juba teguriteks lahutatud, tehke seda.

$$\frac{2\left(x+2y\right)^{2}\left(x^{2}+4y^{2}\right)}{2x\left(x-2y\right)\left(x+2y\right)^{2}\left(x^{2}+4y^{2}\right)}$$

Taandage 2\left(x+2y\right)^{2}\left(x^{2}+4y^{2}\right) nii lugejas kui ka nimetajas.

$$\frac{1}{x\left(x-2y\right)}$$

Laiendage avaldist.

$$\frac{1}{x^{2}-2xy}$$