Solve x/2-frac{1{x}}-1/8x+4+1/5

$$\frac { x } { 2 - \frac { 1 } { x } } - \frac { 1 } { 8 x + 4 } + \frac { 1 } { 5 }$$

Evaluate

$$\frac{40x^{3}+36x^{2}-10x+1}{20\left(4x^{2}-1\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x}{x}.

$$\frac{x}{\frac{2x}{x}-\frac{1}{x}}-\frac{1}{8x+4}+\frac{1}{5}$$

Since \frac{2x}{x} and \frac{1}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x}{\frac{2x-1}{x}}-\frac{1}{8x+4}+\frac{1}{5}$$

Divide x by \frac{2x-1}{x} by multiplying x by the reciprocal of \frac{2x-1}{x}.

$$\frac{xx}{2x-1}-\frac{1}{8x+4}+\frac{1}{5}$$

Multiply x and x to get x^{2}.

$$\frac{x^{2}}{2x-1}-\frac{1}{8x+4}+\frac{1}{5}$$

Factor 8x+4.

$$\frac{x^{2}}{2x-1}-\frac{1}{4\left(2x+1\right)}+\frac{1}{5}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x-1 and 4\left(2x+1\right) is 4\left(2x-1\right)\left(2x+1\right). Multiply \frac{x^{2}}{2x-1} times \frac{4\left(2x+1\right)}{4\left(2x+1\right)}. Multiply \frac{1}{4\left(2x+1\right)} times \frac{2x-1}{2x-1}.

$$\frac{x^{2}\times 4\left(2x+1\right)}{4\left(2x-1\right)\left(2x+1\right)}-\frac{2x-1}{4\left(2x-1\right)\left(2x+1\right)}+\frac{1}{5}$$

Since \frac{x^{2}\times 4\left(2x+1\right)}{4\left(2x-1\right)\left(2x+1\right)} and \frac{2x-1}{4\left(2x-1\right)\left(2x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}\times 4\left(2x+1\right)-\left(2x-1\right)}{4\left(2x-1\right)\left(2x+1\right)}+\frac{1}{5}$$

Do the multiplications in x^{2}\times 4\left(2x+1\right)-\left(2x-1\right).

$$\frac{8x^{3}+4x^{2}-2x+1}{4\left(2x-1\right)\left(2x+1\right)}+\frac{1}{5}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(2x-1\right)\left(2x+1\right) and 5 is 20\left(2x-1\right)\left(2x+1\right). Multiply \frac{8x^{3}+4x^{2}-2x+1}{4\left(2x-1\right)\left(2x+1\right)} times \frac{5}{5}. Multiply \frac{1}{5} times \frac{4\left(2x-1\right)\left(2x+1\right)}{4\left(2x-1\right)\left(2x+1\right)}.

$$\frac{5\left(8x^{3}+4x^{2}-2x+1\right)}{20\left(2x-1\right)\left(2x+1\right)}+\frac{4\left(2x-1\right)\left(2x+1\right)}{20\left(2x-1\right)\left(2x+1\right)}$$

Since \frac{5\left(8x^{3}+4x^{2}-2x+1\right)}{20\left(2x-1\right)\left(2x+1\right)} and \frac{4\left(2x-1\right)\left(2x+1\right)}{20\left(2x-1\right)\left(2x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{5\left(8x^{3}+4x^{2}-2x+1\right)+4\left(2x-1\right)\left(2x+1\right)}{20\left(2x-1\right)\left(2x+1\right)}$$

Do the multiplications in 5\left(8x^{3}+4x^{2}-2x+1\right)+4\left(2x-1\right)\left(2x+1\right).

$$\frac{40x^{3}+20x^{2}-10x+5+16x^{2}+8x-8x-4}{20\left(2x-1\right)\left(2x+1\right)}$$

Combine like terms in 40x^{3}+20x^{2}-10x+5+16x^{2}+8x-8x-4.

$$\frac{40x^{3}+36x^{2}-10x+1}{20\left(2x-1\right)\left(2x+1\right)}$$

Expand 20\left(2x-1\right)\left(2x+1\right).

$$\frac{40x^{3}+36x^{2}-10x+1}{80x^{2}-20}$$

Factor

$$\frac{40x^{3}+36x^{2}-10x+1}{20\left(4x^{2}-1\right)}$$