Solve (1 + sqrt(2))/(2 + sqrt(3)) + (2 + sqrt(3))/(1 + sqrt(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

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

Evaluate

$2^{\frac{5}{2}}-2\sqrt{3}\approx 2.192752634$

Solution Steps

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

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

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

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

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

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

Subtract $3$ from $4$ to get $1$.

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

Anything divided by one gives itself.

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

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

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

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

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

Square $1$. Square $\sqrt{2}$.

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

Subtract $2$ from $1$ to get $-1$.

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

Anything divided by -1 gives its opposite.

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

Apply the distributive property by multiplying each term of $1+\sqrt{2}$ by each term of $2-\sqrt{3}$.

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

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

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

Apply the distributive property by multiplying each term of $2+\sqrt{3}$ by each term of $1-\sqrt{2}$.

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

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

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

To find the opposite of $2-2\sqrt{2}+\sqrt{3}-\sqrt{6}$, find the opposite of each term.

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

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

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

The opposite of $-\sqrt{6}$ is $\sqrt{6}$.

$$2-\sqrt{3}+2\sqrt{2}-\sqrt{6}-2+2\sqrt{2}-\sqrt{3}+\sqrt{6}$$

Subtract $2$ from $2$ to get $0$.

$$-\sqrt{3}+2\sqrt{2}-\sqrt{6}+2\sqrt{2}-\sqrt{3}+\sqrt{6}$$

Combine $2\sqrt{2}$ and $2\sqrt{2}$ to get $4\sqrt{2}$.

$$-\sqrt{3}+4\sqrt{2}-\sqrt{6}-\sqrt{3}+\sqrt{6}$$

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

$$-2\sqrt{3}+4\sqrt{2}-\sqrt{6}+\sqrt{6}$$

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

$$-2\sqrt{3}+4\sqrt{2}$$

Factor

$2 {(2 \sqrt{2} - \sqrt{3})} = 2.192752634$