Solve 1/3^-2+sqrt{3}

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

Evaluate

$$\frac{81\sqrt{3}-9}{242}\approx 0.542545931$$

Solution Steps

Calculate 3 to the power of -2 and get \frac{1}{9}.

$$\frac{1}{\frac{1}{9}+\sqrt{3}}$$

Rationalize the denominator of \frac{1}{\frac{1}{9}+\sqrt{3}} by multiplying numerator and denominator by \frac{1}{9}-\sqrt{3}.

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

Consider \left(\frac{1}{9}+\sqrt{3}\right)\left(\frac{1}{9}-\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}.

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

Square \frac{1}{9}. Square \sqrt{3}.

$$\frac{\frac{1}{9}-\sqrt{3}}{\frac{1}{81}-3}$$

Subtract 3 from \frac{1}{81} to get -\frac{242}{81}.

$$\frac{\frac{1}{9}-\sqrt{3}}{-\frac{242}{81}}$$

Divide \frac{1}{9}-\sqrt{3} by -\frac{242}{81} by multiplying \frac{1}{9}-\sqrt{3} by the reciprocal of -\frac{242}{81}.

$$\frac{\left(\frac{1}{9}-\sqrt{3}\right)\times 81}{-242}$$

Use the distributive property to multiply \frac{1}{9}-\sqrt{3} by 81.

$$\frac{9-81\sqrt{3}}{-242}$$

Factor

$$\frac{9 {(9 \sqrt{3} - 1)}}{242} = 0.5425459314590043$$