Solve 1/1+sqrt{2}+1/sqrt{2+sqrt{3}}+1/sqrt{3+sqrt{4}}=1

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

Verify

Solution Steps

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

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

Consider \left(1+\sqrt{2}\right)\left(1-\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}.

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

Square 1. Square \sqrt{2}.

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

Subtract 2 from 1 to get -1.

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

Anything divided by -1 gives its opposite. To find the opposite of 1-\sqrt{2}, find the opposite of each term.

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

The opposite of -\sqrt{2} is \sqrt{2}.

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

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

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

Consider \left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\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}.

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

Square \sqrt{2}. Square \sqrt{3}.

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

Subtract 3 from 2 to get -1.

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

Anything divided by -1 gives its opposite. To find the opposite of \sqrt{2}-\sqrt{3}, find the opposite of each term.

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

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

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

The opposite of -\sqrt{3} is \sqrt{3}.

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

Calculate the square root of 4 and get 2.

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

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

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

Consider \left(\sqrt{3}+2\right)\left(\sqrt{3}-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}.

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

Square \sqrt{3}. Square 2.

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

Subtract 4 from 3 to get -1.

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

Anything divided by -1 gives its opposite. To find the opposite of \sqrt{3}-2, find the opposite of each term.

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

Combine \sqrt{3} and -\sqrt{3} to get 0.

$$-1-\left(-2\right)=1$$

The opposite of -2 is 2.

$$-1+2=1$$

Add -1 and 2 to get 1.

$$1=1$$

Compare 1 and 1.

$$\text{true}$$