Solve (1/2^2+32+2-1/a^2+42+3)(2^2+52+6)

$$( \frac { 1 } { 2 ^ { 2 } + 32 + 2 } - \frac { 1 } { a ^ { 2 } + 42 + 3 } ) ( 2 ^ { 2 } + 52 + 6 )$$

Evaluate

$$\frac{31\left(a^{2}+7\right)}{19\left(a^{2}+45\right)}$$

Short Solution Steps

Calculate 2 to the power of 2 and get 4.

$$\left(\frac{1}{4+32+2}-\frac{1}{a^{2}+42+3}\right)\left(2^{2}+52+6\right)$$

Add 4 and 32 to get 36.

$$\left(\frac{1}{36+2}-\frac{1}{a^{2}+42+3}\right)\left(2^{2}+52+6\right)$$

Add 36 and 2 to get 38.

$$\left(\frac{1}{38}-\frac{1}{a^{2}+42+3}\right)\left(2^{2}+52+6\right)$$

Add 42 and 3 to get 45.

$$\left(\frac{1}{38}-\frac{1}{a^{2}+45}\right)\left(2^{2}+52+6\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 38 and a^{2}+45 is 38\left(a^{2}+45\right). Multiply \frac{1}{38} times \frac{a^{2}+45}{a^{2}+45}. Multiply \frac{1}{a^{2}+45} times \frac{38}{38}.

$$\left(\frac{a^{2}+45}{38\left(a^{2}+45\right)}-\frac{38}{38\left(a^{2}+45\right)}\right)\left(2^{2}+52+6\right)$$

Since \frac{a^{2}+45}{38\left(a^{2}+45\right)} and \frac{38}{38\left(a^{2}+45\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{a^{2}+45-38}{38\left(a^{2}+45\right)}\left(2^{2}+52+6\right)$$

Combine like terms in a^{2}+45-38.

$$\frac{a^{2}+7}{38\left(a^{2}+45\right)}\left(2^{2}+52+6\right)$$

Calculate 2 to the power of 2 and get 4.

$$\frac{a^{2}+7}{38\left(a^{2}+45\right)}\left(4+52+6\right)$$

Add 4 and 52 to get 56.

$$\frac{a^{2}+7}{38\left(a^{2}+45\right)}\left(56+6\right)$$

Add 56 and 6 to get 62.

$$\frac{a^{2}+7}{38\left(a^{2}+45\right)}\times 62$$

Express \frac{a^{2}+7}{38\left(a^{2}+45\right)}\times 62 as a single fraction.

$$\frac{\left(a^{2}+7\right)\times 62}{38\left(a^{2}+45\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{31\left(a^{2}+7\right)}{19\left(a^{2}+45\right)}$$

Use the distributive property to multiply 31 by a^{2}+7.

$$\frac{31a^{2}+217}{19\left(a^{2}+45\right)}$$

Use the distributive property to multiply 19 by a^{2}+45.

$$\frac{31a^{2}+217}{19a^{2}+855}$$

Expand

$$\frac{31\left(a^{2}+7\right)}{19\left(a^{2}+45\right)}$$

Short Solution Steps

Calculate 2 to the power of 2 and get 4.

$$\left(\frac{1}{4+32+2}-\frac{1}{a^{2}+42+3}\right)\left(2^{2}+52+6\right)$$

Add 4 and 32 to get 36.

$$\left(\frac{1}{36+2}-\frac{1}{a^{2}+42+3}\right)\left(2^{2}+52+6\right)$$

Add 36 and 2 to get 38.

$$\left(\frac{1}{38}-\frac{1}{a^{2}+42+3}\right)\left(2^{2}+52+6\right)$$

Add 42 and 3 to get 45.

$$\left(\frac{1}{38}-\frac{1}{a^{2}+45}\right)\left(2^{2}+52+6\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 38 and a^{2}+45 is 38\left(a^{2}+45\right). Multiply \frac{1}{38} times \frac{a^{2}+45}{a^{2}+45}. Multiply \frac{1}{a^{2}+45} times \frac{38}{38}.

$$\left(\frac{a^{2}+45}{38\left(a^{2}+45\right)}-\frac{38}{38\left(a^{2}+45\right)}\right)\left(2^{2}+52+6\right)$$

Since \frac{a^{2}+45}{38\left(a^{2}+45\right)} and \frac{38}{38\left(a^{2}+45\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{a^{2}+45-38}{38\left(a^{2}+45\right)}\left(2^{2}+52+6\right)$$

Combine like terms in a^{2}+45-38.

$$\frac{a^{2}+7}{38\left(a^{2}+45\right)}\left(2^{2}+52+6\right)$$

Calculate 2 to the power of 2 and get 4.

$$\frac{a^{2}+7}{38\left(a^{2}+45\right)}\left(4+52+6\right)$$

Add 4 and 52 to get 56.

$$\frac{a^{2}+7}{38\left(a^{2}+45\right)}\left(56+6\right)$$

Add 56 and 6 to get 62.

$$\frac{a^{2}+7}{38\left(a^{2}+45\right)}\times 62$$

Express \frac{a^{2}+7}{38\left(a^{2}+45\right)}\times 62 as a single fraction.

$$\frac{\left(a^{2}+7\right)\times 62}{38\left(a^{2}+45\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{31\left(a^{2}+7\right)}{19\left(a^{2}+45\right)}$$

Use the distributive property to multiply 31 by a^{2}+7.

$$\frac{31a^{2}+217}{19\left(a^{2}+45\right)}$$

Use the distributive property to multiply 19 by a^{2}+45.

$$\frac{31a^{2}+217}{19a^{2}+855}$$