Solve (x)/((x)^(2)+x-2)-(3)/((x)^(2)+2x-3)-(x)/((x)^(2)+5x+6) | 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

$$\frac{ x }{ { x }^{ 2 } +x-2 } - \frac{ 3 }{ { x }^{ 2 } +2x-3 } - \frac{ x }{ { x }^{ 2 } +5x+6 }$$

Evaluate

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

Short Solution Steps

Factor $x^{2}+x-2$. Factor $x^{2}+2x-3$.

$$\frac{x}{\left(x-1\right)\left(x+2\right)}-\frac{3}{\left(x-1\right)\left(x+3\right)}-\frac{x}{x^{2}+5x+6}$$

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

$$\frac{x\left(x+3\right)}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}-\frac{3\left(x+2\right)}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}-\frac{x}{x^{2}+5x+6}$$

Since $\frac{x\left(x+3\right)}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}$ and $\frac{3\left(x+2\right)}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x+3\right)-3\left(x+2\right)}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}-\frac{x}{x^{2}+5x+6}$$

Do the multiplications in $x\left(x+3\right)-3\left(x+2\right)$.

$$\frac{x^{2}+3x-3x-6}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}-\frac{x}{x^{2}+5x+6}$$

Combine like terms in $x^{2}+3x-3x-6$.

$$\frac{x^{2}-6}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}-\frac{x}{x^{2}+5x+6}$$

Factor $x^{2}+5x+6$.

$$\frac{x^{2}-6}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}-\frac{x}{\left(x+2\right)\left(x+3\right)}$$

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

$$\frac{x^{2}-6}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}-\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}$$

Since $\frac{x^{2}-6}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}$ and $\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}-6-x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}$$

Do the multiplications in $x^{2}-6-x\left(x-1\right)$.

$$\frac{x^{2}-6-x^{2}+x}{\left(x-1\right)\left(x+2\right)\left(x+3\right)}$$

Combine like terms in $x^{2}-6-x^{2}+x$.

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

Expand $\left(x-1\right)\left(x+2\right)\left(x+3\right)$.

$$\frac{-6+x}{x^{3}+4x^{2}+x-6}$$

Differentiate w.r.t. x

$-\frac{x\left(\left(2x-7\right)^{2}-145\right)}{2\left(\left(x-1\right)\left(x+2\right)\left(x+3\right)\right)^{2}}$