Lutasin ang frac{2sqrt{3}-sqrt{2}}{2sqrt{3}+sqrt{2}}

$$\frac { 2 \sqrt { 3 } - \sqrt { 2 } } { 2 \sqrt { 3 } + \sqrt { 2 } }$$

I-evaluate

$$\frac{7-2\sqrt{6}}{5}\approx 0.420204103$$

Mga Hakbang ng Solution

I-rationalize ang denominator ng \frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{3}+\sqrt{2}} sa pamamagitan ng pag-multiply ng numerator at denominator sa 2\sqrt{3}-\sqrt{2}.

$$\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right)}$$

Isaalang-alang ang \left(2\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right). Maaaring ma-transform ang pag-multiply sa difference ng mga square gamit ang rule na: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

I-multiply ang 2\sqrt{3}-\sqrt{2} at 2\sqrt{3}-\sqrt{2} para makuha ang \left(2\sqrt{3}-\sqrt{2}\right)^{2}.

$$\frac{\left(2\sqrt{3}-\sqrt{2}\right)^{2}}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Gamitin ang binomial theorem na \left(a-b\right)^{2}=a^{2}-2ab+b^{2} para palawakin ang \left(2\sqrt{3}-\sqrt{2}\right)^{2}.

$$\frac{4\left(\sqrt{3}\right)^{2}-4\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Ang square ng \sqrt{3} ay 3.

$$\frac{4\times 3-4\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

I-multiply ang 4 at 3 para makuha ang 12.

$$\frac{12-4\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Para i-multiply ang \sqrt{3} at \sqrt{2}, i-multiply ang mga numero sa ilalim ng square root.

$$\frac{12-4\sqrt{6}+\left(\sqrt{2}\right)^{2}}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Ang square ng \sqrt{2} ay 2.

$$\frac{12-4\sqrt{6}+2}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Idagdag ang 12 at 2 para makuha ang 14.

$$\frac{14-4\sqrt{6}}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Palawakin ang \left(2\sqrt{3}\right)^{2}.

$$\frac{14-4\sqrt{6}}{2^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Kalkulahin ang 2 sa power ng 2 at kunin ang 4.

$$\frac{14-4\sqrt{6}}{4\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Ang square ng \sqrt{3} ay 3.

$$\frac{14-4\sqrt{6}}{4\times 3-\left(\sqrt{2}\right)^{2}}$$

I-multiply ang 4 at 3 para makuha ang 12.

$$\frac{14-4\sqrt{6}}{12-\left(\sqrt{2}\right)^{2}}$$

Ang square ng \sqrt{2} ay 2.

$$\frac{14-4\sqrt{6}}{12-2}$$

I-subtract ang 2 mula sa 12 para makuha ang 10.

$$\frac{14-4\sqrt{6}}{10}$$

I-factor

$$\frac{7 - 2 \sqrt{6}}{5} = 0.4202041028867288$$