Solve (3sqrt(2))/(sqrt(6) + sqrt(3)) + (4sqrt(3))/(sqrt(6) + sqrt(2)) - (4sqrt(2))/(sqrt(6) - sqrt(2)) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{3\sqrt{2}}{\sqrt{6}+\sqrt{3}}+\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}-\frac{4\sqrt{2}}{\sqrt{6}-\sqrt{2}}$$

Evaluate

$3\sqrt{2}-2\sqrt{6}-2\approx -2.656338798$

Short Solution Steps

Rationalize the denominator of $\frac{3\sqrt{2}}{\sqrt{6}+\sqrt{3}}$ by multiplying numerator and denominator by $\sqrt{6}-\sqrt{3}$.

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

Consider $\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

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

Square $\sqrt{6}$. Square $\sqrt{3}$.

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

Subtract $3$ from $6$ to get $3$.

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

Rationalize the denominator of $\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{6}-\sqrt{2}$.

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

Consider $\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

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

Square $\sqrt{6}$. Square $\sqrt{2}$.

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

Subtract $2$ from $6$ to get $4$.

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

Rationalize the denominator of $\frac{4\sqrt{2}}{\sqrt{6}-\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{6}+\sqrt{2}$.

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

Consider $\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

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

Square $\sqrt{6}$. Square $\sqrt{2}$.

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

Subtract $2$ from $6$ to get $4$.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $3$ and $4$ is $12$. Multiply $\frac{3\sqrt{2}\left(\sqrt{6}-\sqrt{3}\right)}{3}$ times $\frac{4}{4}$. Multiply $\frac{4\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right)}{4}$ times $\frac{3}{3}$.

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

Since $\frac{4\times 3\sqrt{2}\left(\sqrt{6}-\sqrt{3}\right)}{12}$ and $\frac{3\times 4\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right)}{12}$ have the same denominator, add them by adding their numerators.

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

Do the multiplications in $4\times 3\sqrt{2}\left(\sqrt{6}-\sqrt{3}\right)+3\times 4\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right)$.

$$\frac{24\sqrt{3}-12\sqrt{6}+36\sqrt{2}-12\sqrt{6}}{12}-\frac{4\sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)}{4}$$

Do the calculations in $24\sqrt{3}-12\sqrt{6}+36\sqrt{2}-12\sqrt{6}$.

$$\frac{24\sqrt{3}-24\sqrt{6}+36\sqrt{2}}{12}-\frac{4\sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)}{4}$$

Divide each term of $24\sqrt{3}-24\sqrt{6}+36\sqrt{2}$ by $12$ to get $2\sqrt{3}-2\sqrt{6}+3\sqrt{2}$.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply $2\sqrt{3}-2\sqrt{6}+3\sqrt{2}$ times $\frac{4}{4}$.

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

Since $\frac{4\left(2\sqrt{3}-2\sqrt{6}+3\sqrt{2}\right)}{4}$ and $\frac{4\sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)}{4}$ have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in $4\left(2\sqrt{3}-2\sqrt{6}+3\sqrt{2}\right)-4\sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)$.

$$\frac{8\sqrt{3}-8\sqrt{6}+12\sqrt{2}-8\sqrt{3}-8}{4}$$

Do the calculations in $8\sqrt{3}-8\sqrt{6}+12\sqrt{2}-8\sqrt{3}-8$.

$$\frac{-8\sqrt{6}+12\sqrt{2}-8}{4}$$

Divide each term of $-8\sqrt{6}+12\sqrt{2}-8$ by $4$ to get $-2\sqrt{6}+3\sqrt{2}-2$.

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