Solve for x 5/(x+1)-1/(1-x)=x/(x-1)

$\frac{5}{x + 1} - \frac{1}{1 - x} = \frac{x}{x - 1}$

Step 1

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x + 1, 1 - x, x - 1$

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factor for $x + 1$ is $x + 1$ itself.

$(x + 1) = x + 1$

$(x + 1)$ occurs $1$ time.

The factor for $1 - x$ is $1 - x$ itself.

$(1 - x) = 1 - x$

$(1 - x)$ occurs $1$ time.

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

The LCM of $x + 1, 1 - x, x - 1$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x + 1) (1 - x) (x - 1)$

Step 2

Multiply each term in $\frac{5}{x + 1} - \frac{1}{1 - x} = \frac{x}{x - 1}$ by $(x + 1) (1 - x) (x - 1)$ to eliminate the fractions.

Multiply each term in $\frac{5}{x + 1} - \frac{1}{1 - x} = \frac{x}{x - 1}$ by $(x + 1) (1 - x) (x - 1)$ .

$\frac{5}{x + 1} ((x + 1) (1 - x) (x - 1)) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Simplify the left side.

Simplify each term.

Cancel the common factor of $x + 1$ .

Factor $x + 1$ out of $(x + 1) (1 - x) (x - 1)$ .

$\frac{5}{x + 1} ((x + 1) ((1 - x) (x - 1))) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Cancel the common factor.

$\frac{5}{x + 1} ((x + 1) ((1 - x) (x - 1))) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Rewrite the expression.

$5 ((1 - x) (x - 1)) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Rewrite $1$ as $- 1 (- 1)$ .

$5 ((- 1 (- 1) - x) (x - 1)) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Factor $- 1$ out of $- x$ .

$5 ((- 1 (- 1) - (x)) (x - 1)) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Factor $- 1$ out of $- 1 (- 1) - (x)$ .

$5 (- 1 (- 1 + x) (x - 1)) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Reorder terms.

$5 (- 1 (x - 1) (x - 1)) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Raise $x - 1$ to the power of $1$ .

$5 (- 1 ({(x - 1)}^{1} (x - 1))) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Raise $x - 1$ to the power of $1$ .

$5 (- 1 ({(x - 1)}^{1} {(x - 1)}^{1})) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$5 (- 1 {(x - 1)}^{1 + 1}) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Add $1$ and $1$ .

$5 (- 1 {(x - 1)}^{2}) - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Multiply $- 1$ by $5$ .

$- 5 {(x - 1)}^{2} - \frac{1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Cancel the common factor of $1 - x$ .

Move the leading negative in $- \frac{1}{1 - x}$ into the numerator.

$- 5 {(x - 1)}^{2} + \frac{- 1}{1 - x} ((x + 1) (1 - x) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Factor $1 - x$ out of $(x + 1) (1 - x) (x - 1)$ .

$- 5 {(x - 1)}^{2} + \frac{- 1}{1 - x} ((1 - x) ((x + 1) (x - 1))) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Cancel the common factor.

$- 5 {(x - 1)}^{2} + \frac{- 1}{1 - x} ((1 - x) ((x + 1) (x - 1))) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Rewrite the expression.

$- 5 {(x - 1)}^{2} - ((x + 1) (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$- 5 {(x - 1)}^{2} - (x (x - 1) + 1 (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Apply the distributive property.

$- 5 {(x - 1)}^{2} - (x \cdot x + x \cdot - 1 + 1 (x - 1)) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Apply the distributive property.

$- 5 {(x - 1)}^{2} - (x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$- 5 {(x - 1)}^{2} - (x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Move $- 1$ to the left of $x$ .

$- 5 {(x - 1)}^{2} - (x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Rewrite $- 1 x$ as $- x$ .

$- 5 {(x - 1)}^{2} - (x^{2} - x + 1 x + 1 \cdot - 1) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Multiply $x$ by $1$ .

$- 5 {(x - 1)}^{2} - (x^{2} - x + x + 1 \cdot - 1) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Multiply $- 1$ by $1$ .

$- 5 {(x - 1)}^{2} - (x^{2} - x + x - 1) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Add $- x$ and $x$ .

$- 5 {(x - 1)}^{2} - (x^{2} + 0 - 1) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Add $x^{2}$ and $0$ .

$- 5 {(x - 1)}^{2} - (x^{2} - 1) = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Apply the distributive property.

$- 5 {(x - 1)}^{2} - x^{2} - - 1 = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Multiply $- 1$ by $- 1$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = \frac{x}{x - 1} ((x + 1) (1 - x) (x - 1))$

Simplify the right side.

Cancel the common factor of $x - 1$ .

Factor $x - 1$ out of $(x + 1) (1 - x) (x - 1)$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = \frac{x}{x - 1} ((x - 1) ((x + 1) (1 - x)))$

Cancel the common factor.

$- 5 {(x - 1)}^{2} - x^{2} + 1 = \frac{x}{x - 1} ((x - 1) ((x + 1) (1 - x)))$

Rewrite the expression.

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x ((x + 1) (1 - x))$

Expand $(x + 1) (1 - x)$ using the FOIL Method.

Apply the distributive property.

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (x (1 - x) + 1 (1 - x))$

Apply the distributive property.

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (x \cdot 1 + x (- x) + 1 (1 - x))$

Apply the distributive property.

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (x \cdot 1 + x (- x) + 1 \cdot 1 + 1 (- x))$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $1$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (x + x (- x) + 1 \cdot 1 + 1 (- x))$

Rewrite using the commutative property of multiplication.

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (x - x \cdot x + 1 \cdot 1 + 1 (- x))$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (x - (x \cdot x) + 1 \cdot 1 + 1 (- x))$

Multiply $x$ by $x$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (x - x^{2} + 1 \cdot 1 + 1 (- x))$

Multiply $1$ by $1$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (x - x^{2} + 1 + 1 (- x))$

Multiply $- x$ by $1$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (x - x^{2} + 1 - x)$

Subtract $x$ from $x$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (- x^{2} + 1 + 0)$

Add $- x^{2}$ and $0$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (- x^{2} + 1)$

Apply the distributive property.

$- 5 {(x - 1)}^{2} - x^{2} + 1 = x (- x^{2}) + x \cdot 1$

Simplify the expression.

Rewrite using the commutative property of multiplication.

$- 5 {(x - 1)}^{2} - x^{2} + 1 = - x \cdot x^{2} + x \cdot 1$

Multiply $x$ by $1$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = - x \cdot x^{2} + x$

Multiply $x$ by $x^{2}$ by adding the exponents.

Move $x^{2}$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = - (x^{2} x) + x$

Multiply $x^{2}$ by $x$ .

Raise $x$ to the power of $1$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = - (x^{2} x^{1}) + x$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 5 {(x - 1)}^{2} - x^{2} + 1 = - x^{2 + 1} + x$

Add $2$ and $1$ .

$- 5 {(x - 1)}^{2} - x^{2} + 1 = - x^{3} + x$

Step 3

Solve the equation.

Move all terms containing $x$ to the left side of the equation.

Add $x^{3}$ to both sides of the equation.

$- 5 {(x - 1)}^{2} - x^{2} + 1 + x^{3} = x$

Subtract $x$ from both sides of the equation.

$- 5 {(x - 1)}^{2} - x^{2} + 1 + x^{3} - x = 0$

Simplify each term.

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$- 5 ((x - 1) (x - 1)) - x^{2} + 1 + x^{3} - x = 0$

Expand $(x - 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$- 5 (x (x - 1) - 1 (x - 1)) - x^{2} + 1 + x^{3} - x = 0$

Apply the distributive property.

$- 5 (x \cdot x + x \cdot - 1 - 1 (x - 1)) - x^{2} + 1 + x^{3} - x = 0$

Apply the distributive property.

$- 5 (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) - x^{2} + 1 + x^{3} - x = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$- 5 (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1) - x^{2} + 1 + x^{3} - x = 0$

Move $- 1$ to the left of $x$ .

$- 5 (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1) - x^{2} + 1 + x^{3} - x = 0$

Rewrite $- 1 x$ as $- x$ .

$- 5 (x^{2} - x - 1 x - 1 \cdot - 1) - x^{2} + 1 + x^{3} - x = 0$

Rewrite $- 1 x$ as $- x$ .

$- 5 (x^{2} - x - x - 1 \cdot - 1) - x^{2} + 1 + x^{3} - x = 0$

Multiply $- 1$ by $- 1$ .

$- 5 (x^{2} - x - x + 1) - x^{2} + 1 + x^{3} - x = 0$

Subtract $x$ from $- x$ .

$- 5 (x^{2} - 2 x + 1) - x^{2} + 1 + x^{3} - x = 0$

Apply the distributive property.

$- 5 x^{2} - 5 (- 2 x) - 5 \cdot 1 - x^{2} + 1 + x^{3} - x = 0$

Simplify.

Multiply $- 2$ by $- 5$ .

$- 5 x^{2} + 10 x - 5 \cdot 1 - x^{2} + 1 + x^{3} - x = 0$

Multiply $- 5$ by $1$ .

$- 5 x^{2} + 10 x - 5 - x^{2} + 1 + x^{3} - x = 0$

Subtract $x^{2}$ from $- 5 x^{2}$ .

$- 6 x^{2} + 10 x - 5 + 1 + x^{3} - x = 0$

Subtract $x$ from $10 x$ .

$- 6 x^{2} + 9 x - 5 + 1 + x^{3} = 0$

Add $- 5$ and $1$ .

$- 6 x^{2} + 9 x - 4 + x^{3} = 0$

Factor the left side of the equation.

Reorder terms.

$x^{3} - 6 x^{2} + 9 x - 4 = 0$

Factor $x^{3} - 6 x^{2} + 9 x - 4$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 4, \pm 2 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 4, \pm 2$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$1^{3} - 6 \cdot 1^{2} + 9 \cdot 1 - 4$

Raise $1$ to the power of $3$ .

$1 - 6 \cdot 1^{2} + 9 \cdot 1 - 4$

Raise $1$ to the power of $2$ .

$1 - 6 \cdot 1 + 9 \cdot 1 - 4$

Multiply $- 6$ by $1$ .

$1 - 6 + 9 \cdot 1 - 4$

Subtract $6$ from $1$ .

$- 5 + 9 \cdot 1 - 4$

Multiply $9$ by $1$ .

$- 5 + 9 - 4$

Add $- 5$ and $9$ .

$4 - 4$

Subtract $4$ from $4$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 6 x^{2} + 9 x - 4}{x - 1}$

Divide $x^{3} - 6 x^{2} + 9 x - 4$ by $x - 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{3}$

$6 x^{2}$

$9 x$

$4$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			+	$x^{3}$	-	$x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - x^{2}$

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$

Divide the highest order term in the dividend $- 5 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$
					-	$5 x^{2}$	+	$5 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x^{2} + 5 x$

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$5 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$5 x$
							+	$4 x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$5 x$
							+	$4 x$	-	$4$

Divide the highest order term in the dividend $4 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$5 x$	+	$4$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$5 x$
							+	$4 x$	-	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$	+	$4$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$5 x$
							+	$4 x$	-	$4$
							+	$4 x$	-	$4$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x - 4$

				$x^{2}$	-	$5 x$	+	$4$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$5 x$
							+	$4 x$	-	$4$
							-	$4 x$	+	$4$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$5 x$	+	$4$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	+	$9 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$5 x$
							+	$4 x$	-	$4$
							-	$4 x$	+	$4$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{2} - 5 x + 4$

Write $x^{3} - 6 x^{2} + 9 x - 4$ as a set of factors.

$(x - 1) (x^{2} - 5 x + 4) = 0$

Factor.

Factor $x^{2} - 5 x + 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $4$ and whose sum is $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

$(x - 1) ((x - 4) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x - 1) (x - 4) (x - 1) = 0$

Combine exponents.

Raise $x - 1$ to the power of $1$ .

$(x - 1) (x - 1) (x - 4) = 0$

Raise $x - 1$ to the power of $1$ .

$(x - 1) (x - 1) (x - 4) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(x - 1)}^{1 + 1} (x - 4) = 0$

Add $1$ and $1$ .

${(x - 1)}^{2} (x - 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(x - 1)}^{2} = 0$

$x - 4 = 0$

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

${(x - 1)}^{2} = 0$

Solve ${(x - 1)}^{2} = 0$ for $x$ .

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make ${(x - 1)}^{2} (x - 4) = 0$ true.

$x = 1, 4$

Step 4

Exclude the solutions that do not make $\frac{5}{x + 1} - \frac{1}{1 - x} = \frac{x}{x - 1}$ true.

$x = 4$

Solve for x 5/(x+1)-1/(1-x)=x/(x-1)

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