Answer :
Sure! Let's simplify each of the given expressions step by step:
### Part (a)
Simplify: [tex]\(\frac{1}{m-n}-\frac{m+n}{m^2-n^2}\)[/tex]
1. The first fraction, [tex]\(\frac{1}{m-n}\)[/tex], is already in its simplest form.
2. For the second fraction, note that [tex]\(m^2 - n^2\)[/tex] can be factored using the difference of squares:
[tex]\[ m^2 - n^2 = (m - n)(m + n) \][/tex]
Thus, the second fraction becomes:
[tex]\[ \frac{m+n}{m^2-n^2} = \frac{m+n}{(m-n)(m+n)} = \frac{m+n}{(m-n)(m+n)} = \frac{1}{m-n} \][/tex]
3. Now combine the two fractions:
[tex]\[ \frac{1}{m-n} - \frac{1}{m-n} = 0 \][/tex]
Thus, the simplified form of part (a) is:
[tex]\[ \boxed{0} \][/tex]
### Part (b)
Simplify: [tex]\(\frac{x+1}{x+2}+\frac{x-2}{x^2-4}\)[/tex]
1. The first fraction, [tex]\(\frac{x+1}{x+2}\)[/tex], is already in its simplest form.
2. For the second fraction, note that [tex]\(x^2 - 4\)[/tex] can be factored using the difference of squares:
[tex]\[ x^2 - 4 = (x - 2)(x + 2) \][/tex]
Thus, the second fraction becomes:
[tex]\[ \frac{x-2}{x^2-4} = \frac{x-2}{(x - 2)(x + 2)} = \frac{1}{x + 2} \][/tex]
3. Now combine the two fractions:
[tex]\[ \frac{x+1}{x+2} + \frac{1}{x+2} = \frac{(x+1) + 1}{x+2} = \frac{x+2}{x+2} = 1 \][/tex]
Thus, the simplified form of part (b) is:
[tex]\[ \boxed{1} \][/tex]
### Part (c)
Simplify: [tex]\(\frac{4x}{x^2-1}-\frac{x+1}{x-1}\)[/tex]
1. The first fraction, [tex]\(\frac{4x}{x^2-1}\)[/tex], can be decomposed using the difference of squares:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
Thus, the first fraction becomes:
[tex]\[ \frac{4x}{(x - 1)(x + 1)} \][/tex]
2. The second fraction, [tex]\(\frac{x+1}{x-1}\)[/tex], is already in its simplest form.
3. To combine these fractions, express them over a common denominator:
[tex]\[ \frac{4x}{(x - 1)(x + 1)} - \frac{x+1}{x-1} = \frac{4x - (x + 1)(x + 1)}{(x - 1)(x + 1)} \][/tex]
4. Simplify the numerator:
[tex]\[ 4x - (x + 1)(x + 1) = 4x - (x^2 + 2x + 1) = 4x - x^2 - 2x - 1 = -x^2 + 2x - 1 \][/tex]
5. Now we write:
[tex]\[ \frac{-x^2 + 2x - 1}{(x - 1)(x + 1)} \][/tex]
6. Simplify:
[tex]\[ \frac{(1 - x)}{x + 1} \][/tex]
Thus, the simplified form of part (c) is:
[tex]\[ \boxed{\frac{1 - x}{x + 1}} \][/tex]
In conclusion:
a) [tex]\(\frac{1}{m-n}-\frac{m+n}{m^2 - n^2}\)[/tex] simplifies to [tex]\(0\)[/tex].
b) [tex]\(\frac{x+1}{x+2}+\frac{x-2}{x^2-4}\)[/tex] simplifies to [tex]\(1\)[/tex].
c) [tex]\(\frac{4x}{x^2-1}-\frac{x+1}{x-1}\)[/tex] simplifies to [tex]\(\frac{1 - x}{x + 1}\)[/tex].
### Part (a)
Simplify: [tex]\(\frac{1}{m-n}-\frac{m+n}{m^2-n^2}\)[/tex]
1. The first fraction, [tex]\(\frac{1}{m-n}\)[/tex], is already in its simplest form.
2. For the second fraction, note that [tex]\(m^2 - n^2\)[/tex] can be factored using the difference of squares:
[tex]\[ m^2 - n^2 = (m - n)(m + n) \][/tex]
Thus, the second fraction becomes:
[tex]\[ \frac{m+n}{m^2-n^2} = \frac{m+n}{(m-n)(m+n)} = \frac{m+n}{(m-n)(m+n)} = \frac{1}{m-n} \][/tex]
3. Now combine the two fractions:
[tex]\[ \frac{1}{m-n} - \frac{1}{m-n} = 0 \][/tex]
Thus, the simplified form of part (a) is:
[tex]\[ \boxed{0} \][/tex]
### Part (b)
Simplify: [tex]\(\frac{x+1}{x+2}+\frac{x-2}{x^2-4}\)[/tex]
1. The first fraction, [tex]\(\frac{x+1}{x+2}\)[/tex], is already in its simplest form.
2. For the second fraction, note that [tex]\(x^2 - 4\)[/tex] can be factored using the difference of squares:
[tex]\[ x^2 - 4 = (x - 2)(x + 2) \][/tex]
Thus, the second fraction becomes:
[tex]\[ \frac{x-2}{x^2-4} = \frac{x-2}{(x - 2)(x + 2)} = \frac{1}{x + 2} \][/tex]
3. Now combine the two fractions:
[tex]\[ \frac{x+1}{x+2} + \frac{1}{x+2} = \frac{(x+1) + 1}{x+2} = \frac{x+2}{x+2} = 1 \][/tex]
Thus, the simplified form of part (b) is:
[tex]\[ \boxed{1} \][/tex]
### Part (c)
Simplify: [tex]\(\frac{4x}{x^2-1}-\frac{x+1}{x-1}\)[/tex]
1. The first fraction, [tex]\(\frac{4x}{x^2-1}\)[/tex], can be decomposed using the difference of squares:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
Thus, the first fraction becomes:
[tex]\[ \frac{4x}{(x - 1)(x + 1)} \][/tex]
2. The second fraction, [tex]\(\frac{x+1}{x-1}\)[/tex], is already in its simplest form.
3. To combine these fractions, express them over a common denominator:
[tex]\[ \frac{4x}{(x - 1)(x + 1)} - \frac{x+1}{x-1} = \frac{4x - (x + 1)(x + 1)}{(x - 1)(x + 1)} \][/tex]
4. Simplify the numerator:
[tex]\[ 4x - (x + 1)(x + 1) = 4x - (x^2 + 2x + 1) = 4x - x^2 - 2x - 1 = -x^2 + 2x - 1 \][/tex]
5. Now we write:
[tex]\[ \frac{-x^2 + 2x - 1}{(x - 1)(x + 1)} \][/tex]
6. Simplify:
[tex]\[ \frac{(1 - x)}{x + 1} \][/tex]
Thus, the simplified form of part (c) is:
[tex]\[ \boxed{\frac{1 - x}{x + 1}} \][/tex]
In conclusion:
a) [tex]\(\frac{1}{m-n}-\frac{m+n}{m^2 - n^2}\)[/tex] simplifies to [tex]\(0\)[/tex].
b) [tex]\(\frac{x+1}{x+2}+\frac{x-2}{x^2-4}\)[/tex] simplifies to [tex]\(1\)[/tex].
c) [tex]\(\frac{4x}{x^2-1}-\frac{x+1}{x-1}\)[/tex] simplifies to [tex]\(\frac{1 - x}{x + 1}\)[/tex].
