Solve t Simplify [(7sqrt(3))/(sqrt(10) + sqrt(3)) - (2sqrt(5))/(sqrt(6) + sqrt(5)) - (3sqrt(2))/(sqrt(15) + 3sqrt(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

$$[ \frac { 7 \sqrt { 3 } } { \sqrt { 10 } + \sqrt { 3 } } - \frac { 2 \sqrt { 5 } } { \sqrt { 6 } + \sqrt { 5 } } - \frac { 3 \sqrt { 2 } } { \sqrt { 15 } + 3 \sqrt { 2 } } ]$$

Evaluate

$\frac{30\sqrt{30}}{3}+\frac{75\sqrt{30}}{3}-\frac{90\sqrt{30}}{3}+6\sqrt{30}+36\sqrt{30}+1-5\sqrt{30}-12\sqrt{30}-30\sqrt{30}\approx 1$

Solution Steps

Rationalize the denominator of $\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}$ by multiplying numerator and denominator by $\sqrt{10}-\sqrt{3}$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{\left(\sqrt{10}+\sqrt{3}\right)\left(\sqrt{10}-\sqrt{3}\right)}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$$

Consider $\left(\sqrt{10}+\sqrt{3}\right)\left(\sqrt{10}-\sqrt{3}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{\left(\sqrt{10}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$$

Square $\sqrt{10}$. Square $\sqrt{3}$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{10-3}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$$

Subtract $3$ from $10$ to get $7$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$$

Rationalize the denominator of $\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}$ by multiplying numerator and denominator by $\sqrt{6}-\sqrt{5}$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$$

Consider $\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{5}\right)^{2}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$$

Square $\sqrt{6}$. Square $\sqrt{5}$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{6-5}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$$

Subtract $5$ from $6$ to get $1$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{1}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$$

Anything divided by one gives itself.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$$

Rationalize the denominator of $\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{15}-3\sqrt{2}$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{\left(\sqrt{15}+3\sqrt{2}\right)\left(\sqrt{15}-3\sqrt{2}\right)}$$

Consider $\left(\sqrt{15}+3\sqrt{2}\right)\left(\sqrt{15}-3\sqrt{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{\left(\sqrt{15}\right)^{2}-\left(3\sqrt{2}\right)^{2}}$$

The square of $\sqrt{15}$ is $15$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-\left(3\sqrt{2}\right)^{2}}$$

Expand $\left(3\sqrt{2}\right)^{2}$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-3^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-9\left(\sqrt{2}\right)^{2}}$$

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

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-9\times 2}$$

Multiply $9$ and $2$ to get $18$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-18}$$

Subtract $18$ from $15$ to get $-3$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{-3}$$

Cancel out $-3$ and $-3$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\left(-\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)\right)$$

The opposite of $-\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$ is $\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$.

$$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

Cancel out $7$ and $7$.

$$\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

Use the distributive property to multiply $\sqrt{3}$ by $\sqrt{10}-\sqrt{3}$.

$$\sqrt{3}\sqrt{10}-\left(\sqrt{3}\right)^{2}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

To multiply $\sqrt{3}$ and $\sqrt{10}$, multiply the numbers under the square root.

$$\sqrt{30}-\left(\sqrt{3}\right)^{2}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

The square of $\sqrt{3}$ is $3$.

$$\sqrt{30}-3-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

Use the distributive property to multiply $2\sqrt{5}$ by $\sqrt{6}-\sqrt{5}$.

$$\sqrt{30}-3-\left(2\sqrt{5}\sqrt{6}-2\left(\sqrt{5}\right)^{2}\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

To multiply $\sqrt{5}$ and $\sqrt{6}$, multiply the numbers under the square root.

$$\sqrt{30}-3-\left(2\sqrt{30}-2\left(\sqrt{5}\right)^{2}\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

The square of $\sqrt{5}$ is $5$.

$$\sqrt{30}-3-\left(2\sqrt{30}-2\times 5\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

Multiply $-2$ and $5$ to get $-10$.

$$\sqrt{30}-3-\left(2\sqrt{30}-10\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

To find the opposite of $2\sqrt{30}-10$, find the opposite of each term.

$$\sqrt{30}-3-2\sqrt{30}-\left(-10\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

The opposite of $-10$ is $10$.

$$\sqrt{30}-3-2\sqrt{30}+10+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

Combine $\sqrt{30}$ and $-2\sqrt{30}$ to get $-\sqrt{30}$.

$$-\sqrt{30}-3+10+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

Add $-3$ and $10$ to get $7$.

$$-\sqrt{30}+7+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)$$

Use the distributive property to multiply $\sqrt{2}$ by $\sqrt{15}-3\sqrt{2}$.

$$-\sqrt{30}+7+\sqrt{2}\sqrt{15}-3\left(\sqrt{2}\right)^{2}$$

To multiply $\sqrt{2}$ and $\sqrt{15}$, multiply the numbers under the square root.

$$-\sqrt{30}+7+\sqrt{30}-3\left(\sqrt{2}\right)^{2}$$

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

$$-\sqrt{30}+7+\sqrt{30}-3\times 2$$

Multiply $-3$ and $2$ to get $-6$.

$$-\sqrt{30}+7+\sqrt{30}-6$$

Combine $-\sqrt{30}$ and $\sqrt{30}$ to get $0$.

$$7-6$$

Subtract $6$ from $7$ to get $1$.

$$1$$