Lutasin ang frac{4sqrt{3}}{2-sqrt{2}}-30/4sqrt{3-sqrt{18}}-frac{sqrt{18}}{3-sqrt{12}}

$$\frac { 4 \sqrt { 3 } } { 2 - \sqrt { 2 } } - \frac { 30 } { 4 \sqrt { 3 } - \sqrt { 18 } } - \frac { \sqrt { 18 } } { 3 - \sqrt { 12 } }$$

I-evaluate

$$4\sqrt{6}\approx 9.797958971$$

Maiikling Hakbang sa Solution

I-rationalize ang denominator ng \frac{4\sqrt{3}}{2-\sqrt{2}} sa pamamagitan ng pag-multiply ng numerator at denominator sa 2+\sqrt{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Isaalang-alang ang \left(2-\sqrt{2}\right)\left(2+\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}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2^{2}-\left(\sqrt{2}\right)^{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

I-square ang 2. I-square ang \sqrt{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{4-2}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

I-subtract ang 2 mula sa 4 para makuha ang 2.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

I-factor out ang 18=3^{2}\times 2. I-rewrite ang square root ng product na \sqrt{3^{2}\times 2} bilang product ng mga square root na \sqrt{3^{2}}\sqrt{2}. Kunin ang square root ng 3^{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30}{4\sqrt{3}-3\sqrt{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

I-rationalize ang denominator ng \frac{30}{4\sqrt{3}-3\sqrt{2}} sa pamamagitan ng pag-multiply ng numerator at denominator sa 4\sqrt{3}+3\sqrt{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{\left(4\sqrt{3}-3\sqrt{2}\right)\left(4\sqrt{3}+3\sqrt{2}\right)}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Isaalang-alang ang \left(4\sqrt{3}-3\sqrt{2}\right)\left(4\sqrt{3}+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}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{\left(4\sqrt{3}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Palawakin ang \left(4\sqrt{3}\right)^{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{4^{2}\left(\sqrt{3}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Kalkulahin ang 4 sa power ng 2 at kunin ang 16.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{16\left(\sqrt{3}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Ang square ng \sqrt{3} ay 3.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{16\times 3-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

I-multiply ang 16 at 3 para makuha ang 48.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

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

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-\left(-3\right)^{2}\left(\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Kalkulahin ang -3 sa power ng 2 at kunin ang 9.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-9\left(\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Ang square ng \sqrt{2} ay 2.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-9\times 2}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

I-multiply ang 9 at 2 para makuha ang 18.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-18}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

I-subtract ang 18 mula sa 48 para makuha ang 30.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{30}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

I-cancel out ang 30 at 30.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\left(4\sqrt{3}+3\sqrt{2}\right)-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Para hanapin ang kabaligtaran ng 4\sqrt{3}+3\sqrt{2}, hanapin ang kabaligtaran ng bawat term.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-4\sqrt{3}-3\sqrt{2}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

I-factor out ang 18=3^{2}\times 2. I-rewrite ang square root ng product na \sqrt{3^{2}\times 2} bilang product ng mga square root na \sqrt{3^{2}}\sqrt{2}. Kunin ang square root ng 3^{2}.

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

I-factor out ang 12=2^{2}\times 3. I-rewrite ang square root ng product na \sqrt{2^{2}\times 3} bilang product ng mga square root na \sqrt{2^{2}}\sqrt{3}. Kunin ang square root ng 2^{2}.

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

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

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

Isaalang-alang ang \left(3-2\sqrt{3}\right)\left(3+2\sqrt{3}\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}.

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

Kalkulahin ang 3 sa power ng 2 at kunin ang 9.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-4\sqrt{3}-3\sqrt{2}-\frac{3\sqrt{2}\left(3+2\sqrt{3}\right)}{9-\left(-2\sqrt{3}\right)^{2}}$$

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

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-4\sqrt{3}-3\sqrt{2}-\frac{3\sqrt{2}\left(3+2\sqrt{3}\right)}{9-\left(-2\right)^{2}\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-4\sqrt{3}-3\sqrt{2}-\frac{3\sqrt{2}\left(3+2\sqrt{3}\right)}{9-4\left(\sqrt{3}\right)^{2}}$$

Ang square ng \sqrt{3} ay 3.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-4\sqrt{3}-3\sqrt{2}-\frac{3\sqrt{2}\left(3+2\sqrt{3}\right)}{9-4\times 3}$$

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

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-4\sqrt{3}-3\sqrt{2}-\frac{3\sqrt{2}\left(3+2\sqrt{3}\right)}{9-12}$$

I-subtract ang 12 mula sa 9 para makuha ang -3.

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

I-cancel out ang -3 at -3.

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

Ang kabaliktaran ng -\sqrt{2}\left(3+2\sqrt{3}\right) ay \sqrt{2}\left(3+2\sqrt{3}\right).

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

Para magdagdag o mag-subtract ng mga expression, i-expand ang mga iyon para gawing magkakapareho ang mga denominator ng mga ito. I-multiply ang -4\sqrt{3}-3\sqrt{2} times \frac{2}{2}.

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

Dahil may parehong denominator ang \frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2} at \frac{2\left(-4\sqrt{3}-3\sqrt{2}\right)}{2}, pagsamahin ang mga ito sa pamamagitan ng pagsasama sa mga numerator ng mga ito.

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

Gawin ang mga pag-multiply sa 4\sqrt{3}\left(2+\sqrt{2}\right)+2\left(-4\sqrt{3}-3\sqrt{2}\right).

$$\frac{8\sqrt{3}+4\sqrt{6}-8\sqrt{3}-6\sqrt{2}}{2}+\sqrt{2}\left(3+2\sqrt{3}\right)$$

Kalkulahin ang 8\sqrt{3}+4\sqrt{6}-8\sqrt{3}-6\sqrt{2}.

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

Hati-hatiin ang bawat termino ng 4\sqrt{6}-6\sqrt{2} sa 2 para makuha ang 2\sqrt{6}-3\sqrt{2}.

$$2\sqrt{6}-3\sqrt{2}+\sqrt{2}\left(3+2\sqrt{3}\right)$$

Gamitin ang distributive property para i-multiply ang \sqrt{2} gamit ang 3+2\sqrt{3}.

$$2\sqrt{6}-3\sqrt{2}+3\sqrt{2}+2\sqrt{2}\sqrt{3}$$

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

$$2\sqrt{6}-3\sqrt{2}+3\sqrt{2}+2\sqrt{6}$$

Pagsamahin ang -3\sqrt{2} at 3\sqrt{2} para makuha ang 0.

$$2\sqrt{6}+2\sqrt{6}$$

Pagsamahin ang 2\sqrt{6} at 2\sqrt{6} para makuha ang 4\sqrt{6}.

$$4\sqrt{6}$$