Solve A=2/sqrt{3+1}+2/sqrt{6+sqrt{3}}+2/sqrt{7+sqrt{5}}

$$A = \frac { 2 } { \sqrt { 3 } + 1 } + \frac { 2 } { \sqrt { 6 } + \sqrt { 3 } } + \frac { 2 } { \sqrt { 7 } + \sqrt { 5 } }$$

Solve for A

$$A=\frac{\sqrt{3}}{3}+\frac{2\sqrt{6}}{3}+\sqrt{7}-\sqrt{5}-1\approx 1.620026765$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{2}{\sqrt{3}+1} by multiplying numerator and denominator by \sqrt{3}-1.

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

Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-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}.

$$A=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^{2}-1^{2}}+\frac{2}{\sqrt{6}+\sqrt{3}}+\frac{2}{\sqrt{7}+\sqrt{5}}$$

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Cancel out 2 and 2.

$$A=\sqrt{3}-1+\frac{2}{\sqrt{6}+\sqrt{3}}+\frac{2}{\sqrt{7}+\sqrt{5}}$$

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

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

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}.

$$A=\sqrt{3}-1+\frac{2\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{3}\right)^{2}}+\frac{2}{\sqrt{7}+\sqrt{5}}$$

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

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

Subtract 3 from 6 to get 3.

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

Rationalize the denominator of \frac{2}{\sqrt{7}+\sqrt{5}} by multiplying numerator and denominator by \sqrt{7}-\sqrt{5}.

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

Consider \left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\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}.

$$A=\sqrt{3}-1+\frac{2\left(\sqrt{6}-\sqrt{3}\right)}{3}+\frac{2\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{5}\right)^{2}}$$

Square \sqrt{7}. Square \sqrt{5}.

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

Subtract 5 from 7 to get 2.

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

Cancel out 2 and 2.

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

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

$$A=\sqrt{3}-1+\frac{2\sqrt{6}-2\sqrt{3}}{3}+\sqrt{7}-\sqrt{5}$$

Divide each term of 2\sqrt{6}-2\sqrt{3} by 3 to get \frac{2}{3}\sqrt{6}-\frac{2}{3}\sqrt{3}.

$$A=\sqrt{3}-1+\frac{2}{3}\sqrt{6}-\frac{2}{3}\sqrt{3}+\sqrt{7}-\sqrt{5}$$

Combine \sqrt{3} and -\frac{2}{3}\sqrt{3} to get \frac{1}{3}\sqrt{3}.

$$A=\frac{1}{3}\sqrt{3}-1+\frac{2}{3}\sqrt{6}+\sqrt{7}-\sqrt{5}$$

Assign A

$$A≔\frac{\sqrt{3}}{3}+\frac{2\sqrt{6}}{3}+\sqrt{7}-\sqrt{5}-1$$