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Simplify the following algebraic expressions: 1. (1)/(x-y) - (1)/(x+y) + (2x)/(x2-y2) 2. (x+2)/(1+x+x2) - (x-2)/(1-x+x2) - (2x)/(1+x2+x4)

Mathematics

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Simplify the expression:

1x−y−1x+y+2xx2−y2\frac{1}{x-y} - \frac{1}{x+y} + \frac{2x}{x^2-y^2}

  1. Find a common denominator.

    We know that x2−y2=(x−y)(x+y)x^2-y^2 = (x-y)(x+y). So, the common denominator for all terms is x2−y2x^2-y^2.

    1x−y=1⋅(x+y)(x−y)(x+y)=x+yx2−y2\frac{1}{x-y} = \frac{1 \cdot (x+y)}{(x-y)(x+y)} = \frac{x+y}{x^2-y^2} 1x+y=1⋅(x−y)(x+y)(x−y)=x−yx2−y2\frac{1}{x+y} = \frac{1 \cdot (x-y)}{(x+y)(x-y)} = \frac{x-y}{x^2-y^2}

    Substitute these into the original expression:

    x+yx2−y2−x−yx2−y2+2xx2−y2\frac{x+y}{x^2-y^2} - \frac{x-y}{x^2-y^2} + \frac{2x}{x^2-y^2}

  2. Combine the numerators over the common denominator.

    (x+y)−(x−y)+2xx2−y2\frac{(x+y) - (x-y) + 2x}{x^2-y^2}

  3. Simplify the numerator.

    x+y−x+y+2xx2−y2\frac{x+y-x+y+2x}{x^2-y^2} 2y+2xx2−y2\frac{2y+2x}{x^2-y^2}

  4. Factor the numerator and simplify further.

    2(y+x)x2−y2\frac{2(y+x)}{x^2-y^2}

    Since x2−y2=(x−y)(x+y)x^2-y^2 = (x-y)(x+y), we can write:

    2(x+y)(x−y)(x+y)\frac{2(x+y)}{(x-y)(x+y)}

    Assuming x+y≠0x+y \neq 0, we can cancel the (x+y)(x+y) term.

    2x−y\boxed{\frac{2}{x-y}}

    :
    Simplify the expression:

    x+21+x+x2−x−21−x+x2−2x1+x2+x4\frac{x+2}{1+x+x^2} - \frac{x-2}{1-x+x^2} - \frac{2x}{1+x^2+x^4}

  5. Identify the relationship between the denominators.

    Notice that (1+x+x2)(1−x+x2)=((1+x2)+x)((1+x2)−x)=(1+x2)2−x2=1+2x2+x4−x2=1+x2+x4(1+x+x^2)(1-x+x^2) = ( (1+x^2)+x ) ( (1+x^2)-x ) = (1+x^2)^2 - x^2 = 1+2x^2+x^4-x^2 = 1+x^2+x^4.
    This means the common denominator for the first two terms is the denominator of the third term.

  6. Combine the first two fractions using the common denominator 1+x2+x41+x^2+x^4.

    (x+2)(1−x+x2)(1+x+x2)(1−x+x2)−(x−2)(1+x+x2)(1−x+x2)(1+x+x2)\frac{(x+2)(1-x+x^2)}{(1+x+x^2)(1-x+x^2)} - \frac{(x-2)(1+x+x^2)}{(1-x+x^2)(1+x+x^2)} (x+2)(1−x+x2)−(x−2)(1+x+x2)1+x2+x4\frac{(x+2)(1-x+x^2) - (x-2)(1+x+x^2)}{1+x^2+x^4}

  7. Expand the numerators.

    First part: (x+2)(1−x+x2)=x(1−x+x2)+2(1−x+x2)=x−x2+x3+2−2x+2x2=x3+x2−x+2(x+2)(1-x+x^2) = x(1-x+x^2) + 2(1-x+x^2) = x-x^2+x^3+2-2x+2x^2 = x^3+x^2-x+2.
    Second part: (x−2)(1+x+x2)=x(1+x+x2)−2(1+x+x2)=x+x2+x3−2−2x−2x2=x3−x2−x−2(x-2)(1+x+x^2) = x(1+x+x^2) - 2(1+x+x^2) = x+x^2+x^3-2-2x-2x^2 = x^3-x^2-x-2.

  8. Substitute the expanded numerators back and simplify.

    (x3+x2−x+2)−(x3−x2−x−2)1+x2+x4\frac{(x^3+x^2-x+2) - (x^3-x^2-x-2)}{1+x^2+x^4} x3+x2−x+2−x3+x2+x+21+x2+x4\frac{x^3+x^2-x+2-x^3+x^2+x+2}{1+x^2+x^4} 2x2+41+x2+x4\frac{2x^2+4}{1+x^2+x^4}

  9. Combine this result with the third term of the original expression.

    2x2+41+x2+x4−2x1+x2+x4\frac{2x^2+4}{1+x^2+x^4} - \frac{2x}{1+x^2+x^4} 2x2+4−2x1+x2+x4\frac{2x^2+4-2x}{1+x^2+x^4}

  10. Rearrange the numerator and factor out 2.

    2(x2−x+2)1+x2+x4\boxed{\frac{2(x^2-x+2)}{1+x^2+x^4}}

    :
    Simplify the expression:

    1b+c+b+cb2−c2\frac{1}{b+c} + \frac{b+c}{b^2-c^2}

  11. Find a common denominator.

    We know that b2−c2=(b−c)(b+c)b^2-c^2 = (b-c)(b+c). So, the common denominator is b2−c2b^2-c^2.

    1b+c=1⋅(b−c)(b+c)(b−c)=b−cb2−c2\frac{1}{b+c} = \frac{1 \cdot (b-c)}{(b+c)(b-c)} = \frac{b-c}{b^2-c^2}

    Substitute this into the original expression:

    b−cb2−c2+b+cb2−c2\frac{b-c}{b^2-c^2} + \frac{b+c}{b^2-c^2}

  12. Combine the numerators over the common denominator.

    (b−c)+(b+c)b2−c2\frac{(b-c) + (b+c)}{b^2-c^2}

  13. Simplify the numerator.

    b−c+b+cb2−c2\frac{b-c+b+c}{b^2-c^2} 2bb2−c2\frac{2b}{b^2-c^2} 2bb2−c2\boxed{\frac{2b}{b^2-c^2}}
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