Solve 10/sqrt{6+1}+6/3+sqrt{6}

$$\frac { 10 } { \sqrt { 6 } + 1 } + \frac { 6 } { 3 + \sqrt { 6 } }$$

Evaluate

$$4$$

Solution Steps

Rationalize the denominator of \frac{10}{\sqrt{6}+1} by multiplying numerator and denominator by \sqrt{6}-1.

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

Consider \left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{6}. Square 1.

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

Subtract 1 from 6 to get 5.

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

Divide 10\left(\sqrt{6}-1\right) by 5 to get 2\left(\sqrt{6}-1\right).

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

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

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

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

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

Square 3. Square \sqrt{6}.

$$2\left(\sqrt{6}-1\right)+\frac{6\left(3-\sqrt{6}\right)}{9-6}$$

Subtract 6 from 9 to get 3.

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

Divide 6\left(3-\sqrt{6}\right) by 3 to get 2\left(3-\sqrt{6}\right).

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

Use the distributive property to multiply 2 by \sqrt{6}-1.

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

Use the distributive property to multiply 2 by 3-\sqrt{6}.

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

Add -2 and 6 to get 4.

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

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

$$4$$

Factor

$$2^{2}$$