Solve 5x/2x^2-x-1=4x-5/x^2-1+5/2x+1

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

Solve for x

$$x=-\frac{5}{8}=-0.625$$

$$x=2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -1,-\frac{1}{2},1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right)\left(2x+1\right), the least common multiple of 2x^{2}-x-1,x^{2}-1,2x+1.

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

Use the distributive property to multiply x+1 by 5.

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

Use the distributive property to multiply 5x+5 by x.

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

Use the distributive property to multiply 2x+1 by 4x-5 and combine like terms.

$$5x^{2}+5x=8x^{2}-6x-5+\left(x^{2}-1\right)\times 5$$

Use the distributive property to multiply x^{2}-1 by 5.

$$5x^{2}+5x=8x^{2}-6x-5+5x^{2}-5$$

Combine 8x^{2} and 5x^{2} to get 13x^{2}.

$$5x^{2}+5x=13x^{2}-6x-5-5$$

Subtract 5 from -5 to get -10.

$$5x^{2}+5x=13x^{2}-6x-10$$

Subtract 13x^{2} from both sides.

$$5x^{2}+5x-13x^{2}=-6x-10$$

Combine 5x^{2} and -13x^{2} to get -8x^{2}.

$$-8x^{2}+5x=-6x-10$$

Add 6x to both sides.

$$-8x^{2}+5x+6x=-10$$

Combine 5x and 6x to get 11x.

$$-8x^{2}+11x=-10$$

Add 10 to both sides.

$$-8x^{2}+11x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -8x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=11 ab=-8\times 10=-80$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 11.

$$a=16 b=-5$$

Rewrite -8x^{2}+11x+10 as \left(-8x^{2}+16x\right)+\left(-5x+10\right).

$$\left(-8x^{2}+16x\right)+\left(-5x+10\right)$$

Factor out 8x in the first and 5 in the second group.

$$8x\left(-x+2\right)+5\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(8x+5\right)$$

To find equation solutions, solve -x+2=0 and 8x+5=0.

$$x=2 x=-\frac{5}{8}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,-\frac{1}{2},1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right)\left(2x+1\right), the least common multiple of 2x^{2}-x-1,x^{2}-1,2x+1.

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

Use the distributive property to multiply x+1 by 5.

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

Use the distributive property to multiply 5x+5 by x.

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

Use the distributive property to multiply 2x+1 by 4x-5 and combine like terms.

$$5x^{2}+5x=8x^{2}-6x-5+\left(x^{2}-1\right)\times 5$$

Use the distributive property to multiply x^{2}-1 by 5.

$$5x^{2}+5x=8x^{2}-6x-5+5x^{2}-5$$

Combine 8x^{2} and 5x^{2} to get 13x^{2}.

$$5x^{2}+5x=13x^{2}-6x-5-5$$

Subtract 5 from -5 to get -10.

$$5x^{2}+5x=13x^{2}-6x-10$$

Subtract 13x^{2} from both sides.

$$5x^{2}+5x-13x^{2}=-6x-10$$

Combine 5x^{2} and -13x^{2} to get -8x^{2}.

$$-8x^{2}+5x=-6x-10$$

Add 6x to both sides.

$$-8x^{2}+5x+6x=-10$$

Combine 5x and 6x to get 11x.

$$-8x^{2}+11x=-10$$

Add 10 to both sides.

$$-8x^{2}+11x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 11 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-8\right)\times 10}}{2\left(-8\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-8\right)\times 10}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-11±\sqrt{121+32\times 10}}{2\left(-8\right)}$$

Multiply 32 times 10.

$$x=\frac{-11±\sqrt{121+320}}{2\left(-8\right)}$$

Add 121 to 320.

$$x=\frac{-11±\sqrt{441}}{2\left(-8\right)}$$

Take the square root of 441.

$$x=\frac{-11±21}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-11±21}{-16}$$

Now solve the equation x=\frac{-11±21}{-16} when ± is plus. Add -11 to 21.

$$x=\frac{10}{-16}$$

Reduce the fraction \frac{10}{-16} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{8}$$

Now solve the equation x=\frac{-11±21}{-16} when ± is minus. Subtract 21 from -11.

$$x=-\frac{32}{-16}$$

Divide -32 by -16.

$$x=2$$

The equation is now solved.

$$x=-\frac{5}{8} x=2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,-\frac{1}{2},1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right)\left(2x+1\right), the least common multiple of 2x^{2}-x-1,x^{2}-1,2x+1.

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

Use the distributive property to multiply x+1 by 5.

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

Use the distributive property to multiply 5x+5 by x.

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

Use the distributive property to multiply 2x+1 by 4x-5 and combine like terms.

$$5x^{2}+5x=8x^{2}-6x-5+\left(x^{2}-1\right)\times 5$$

Use the distributive property to multiply x^{2}-1 by 5.

$$5x^{2}+5x=8x^{2}-6x-5+5x^{2}-5$$

Combine 8x^{2} and 5x^{2} to get 13x^{2}.

$$5x^{2}+5x=13x^{2}-6x-5-5$$

Subtract 5 from -5 to get -10.

$$5x^{2}+5x=13x^{2}-6x-10$$

Subtract 13x^{2} from both sides.

$$5x^{2}+5x-13x^{2}=-6x-10$$

Combine 5x^{2} and -13x^{2} to get -8x^{2}.

$$-8x^{2}+5x=-6x-10$$

Add 6x to both sides.

$$-8x^{2}+5x+6x=-10$$

Combine 5x and 6x to get 11x.

$$-8x^{2}+11x=-10$$

Divide both sides by -8.

$$\frac{-8x^{2}+11x}{-8}=-\frac{10}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{11}{-8}x=-\frac{10}{-8}$$

Divide 11 by -8.

$$x^{2}-\frac{11}{8}x=-\frac{10}{-8}$$

Reduce the fraction \frac{-10}{-8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{8}x=\frac{5}{4}$$

Divide -\frac{11}{8}, the coefficient of the x term, by 2 to get -\frac{11}{16}. Then add the square of -\frac{11}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{8}x+\left(-\frac{11}{16}\right)^{2}=\frac{5}{4}+\left(-\frac{11}{16}\right)^{2}$$

Square -\frac{11}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{8}x+\frac{121}{256}=\frac{5}{4}+\frac{121}{256}$$

Add \frac{5}{4} to \frac{121}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{8}x+\frac{121}{256}=\frac{441}{256}$$

Factor x^{2}-\frac{11}{8}x+\frac{121}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{16}\right)^{2}=\frac{441}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{16}\right)^{2}}=\sqrt{\frac{441}{256}}$$

Simplify.

$$x-\frac{11}{16}=\frac{21}{16} x-\frac{11}{16}=-\frac{21}{16}$$

Add \frac{11}{16} to both sides of the equation.

$$x=2 x=-\frac{5}{8}$$