Solve 1/(4sqrt(2)) * (1 + sqrt(3)) - (1/(4sqrt(2)) * (3 + sqrt(3))) + 1/(4sqrt(2)) * (3 - sqrt(3)) - (1/(4sqrt(2)) * (1 - sqrt(3))) = | 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

$$\frac{1}{4\sqrt{2}}(1+\sqrt{3})-(\frac{1}{4\sqrt{2}}(3+\sqrt{3}))+\frac{1}{4\sqrt{2}}(3-\sqrt{3})-(\frac{1}{4\sqrt{2}}(1-\sqrt{3}))=$$

Evaluate

$0$

Short Solution Steps

Rationalize the denominator of $\frac{1}{4\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{2}$.

$$\frac{\sqrt{2}}{4\left(\sqrt{2}\right)^{2}}\left(1+\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(3+\sqrt{3}\right)+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

The square of $\sqrt{2}$ is $2$.

$$\frac{\sqrt{2}}{4\times 2}\left(1+\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(3+\sqrt{3}\right)+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Multiply $4$ and $2$ to get $8$.

$$\frac{\sqrt{2}}{8}\left(1+\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(3+\sqrt{3}\right)+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Express $\frac{\sqrt{2}}{8}\left(1+\sqrt{3}\right)$ as a single fraction.

$$\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{8}-\frac{1}{4\sqrt{2}}\left(3+\sqrt{3}\right)+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Rationalize the denominator of $\frac{1}{4\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{2}$.

$$\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{8}-\frac{\sqrt{2}}{4\left(\sqrt{2}\right)^{2}}\left(3+\sqrt{3}\right)+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

The square of $\sqrt{2}$ is $2$.

$$\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{8}-\frac{\sqrt{2}}{4\times 2}\left(3+\sqrt{3}\right)+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Multiply $4$ and $2$ to get $8$.

$$\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{8}-\frac{\sqrt{2}}{8}\left(3+\sqrt{3}\right)+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Express $\frac{\sqrt{2}}{8}\left(3+\sqrt{3}\right)$ as a single fraction.

$$\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{8}-\frac{\sqrt{2}\left(3+\sqrt{3}\right)}{8}+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Use the distributive property to multiply $\sqrt{2}$ by $3+\sqrt{3}$.

$$\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{8}-\frac{3\sqrt{2}+\sqrt{2}\sqrt{3}}{8}+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

To multiply $\sqrt{2}$ and $\sqrt{3}$, multiply the numbers under the square root.

$$\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{8}-\frac{3\sqrt{2}+\sqrt{6}}{8}+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Since $\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{8}$ and $\frac{3\sqrt{2}+\sqrt{6}}{8}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{\sqrt{2}\left(1+\sqrt{3}\right)-\left(3\sqrt{2}+\sqrt{6}\right)}{8}+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Do the multiplications in $\sqrt{2}\left(1+\sqrt{3}\right)-\left(3\sqrt{2}+\sqrt{6}\right)$.

$$\frac{\sqrt{2}+\sqrt{6}-3\sqrt{2}-\sqrt{6}}{8}+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

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

$$\frac{-2\sqrt{2}}{8}+\frac{1}{4\sqrt{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Rationalize the denominator of $\frac{1}{4\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{2}$.

$$\frac{-2\sqrt{2}}{8}+\frac{\sqrt{2}}{4\left(\sqrt{2}\right)^{2}}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

The square of $\sqrt{2}$ is $2$.

$$\frac{-2\sqrt{2}}{8}+\frac{\sqrt{2}}{4\times 2}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Multiply $4$ and $2$ to get $8$.

$$\frac{-2\sqrt{2}}{8}+\frac{\sqrt{2}}{8}\left(3-\sqrt{3}\right)-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Express $\frac{\sqrt{2}}{8}\left(3-\sqrt{3}\right)$ as a single fraction.

$$\frac{-2\sqrt{2}}{8}+\frac{\sqrt{2}\left(3-\sqrt{3}\right)}{8}-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Since $\frac{-2\sqrt{2}}{8}$ and $\frac{\sqrt{2}\left(3-\sqrt{3}\right)}{8}$ have the same denominator, add them by adding their numerators.

$$\frac{-2\sqrt{2}+\sqrt{2}\left(3-\sqrt{3}\right)}{8}-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

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

$$\frac{-2\sqrt{2}+3\sqrt{2}-\sqrt{6}}{8}-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

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

$$\frac{\sqrt{2}-\sqrt{6}}{8}-\frac{1}{4\sqrt{2}}\left(1-\sqrt{3}\right)$$

Rationalize the denominator of $\frac{1}{4\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{2}$.

$$\frac{\sqrt{2}-\sqrt{6}}{8}-\frac{\sqrt{2}}{4\left(\sqrt{2}\right)^{2}}\left(1-\sqrt{3}\right)$$

The square of $\sqrt{2}$ is $2$.

$$\frac{\sqrt{2}-\sqrt{6}}{8}-\frac{\sqrt{2}}{4\times 2}\left(1-\sqrt{3}\right)$$

Multiply $4$ and $2$ to get $8$.

$$\frac{\sqrt{2}-\sqrt{6}}{8}-\frac{\sqrt{2}}{8}\left(1-\sqrt{3}\right)$$

Express $\frac{\sqrt{2}}{8}\left(1-\sqrt{3}\right)$ as a single fraction.

$$\frac{\sqrt{2}-\sqrt{6}}{8}-\frac{\sqrt{2}\left(1-\sqrt{3}\right)}{8}$$

Use the distributive property to multiply $\sqrt{2}$ by $1-\sqrt{3}$.

$$\frac{\sqrt{2}-\sqrt{6}}{8}-\frac{\sqrt{2}-\sqrt{2}\sqrt{3}}{8}$$

To multiply $\sqrt{2}$ and $\sqrt{3}$, multiply the numbers under the square root.

$$\frac{\sqrt{2}-\sqrt{6}}{8}-\frac{\sqrt{2}-\sqrt{6}}{8}$$

Combine $\frac{\sqrt{2}-\sqrt{6}}{8}$ and $-\frac{\sqrt{2}-\sqrt{6}}{8}$ to get $0$.

$$0$$

Factor

$0$