Answer :
Certainly! Let's go through each expression one by one and simplify them step-by-step.
### Part (d)
Simplify the expression:
[tex]\[ \frac{a^2 + 1}{a + 1} + \frac{a^2 - a + 1}{a^2 - 1} \][/tex]
#### Step 1: Simplify [tex]\(\frac{a^2 - a + 1}{a^2 - 1}\)[/tex]
Notice that [tex]\(a^2 - 1\)[/tex] is a difference of squares:
[tex]\[ a^2 - 1 = (a - 1)(a + 1) \][/tex]
Therefore, we have:
[tex]\[ \frac{a^2 - a + 1}{(a - 1)(a + 1)} \][/tex]
#### Step 2: Combine the fractions with a common denominator
The common denominator for the two fractions is [tex]\((a + 1)(a - 1)\)[/tex].
[tex]\[ \frac{a^2 + 1}{a + 1} = \frac{(a^2 + 1)(a - 1)}{(a + 1)(a - 1)} \][/tex]
This gives us:
[tex]\[ \frac{(a^2 + 1)(a - 1) + (a^2 - a + 1)}{(a + 1)(a - 1)} \][/tex]
#### Step 3: Simplify the numerator
Expand [tex]\( (a^2 + 1)(a - 1) \)[/tex]:
[tex]\[ (a^2 + 1)(a - 1) = a^3 - a^2 + a - 1 \][/tex]
Add this result to [tex]\(a^2 - a + 1\)[/tex]:
[tex]\[ a^3 - a^2 + a - 1 + a^2 - a + 1 \][/tex]
Combine like terms:
[tex]\[ a^3 - a^2 + a^2 + a - a - 1 + 1 = a^3 \][/tex]
Thus, we have:
[tex]\[ \frac{a^3}{(a + 1)(a - 1)} = \frac{a^3}{a^2 - 1} \][/tex]
Since [tex]\(a^3 = a \cdot (a^2)\)[/tex], this fraction simplifies to:
[tex]\[ a \ \text{(as the terms cancel out)} \][/tex]
Hence,
[tex]\[ \frac{a^2+1}{a+1}+\frac{a^2-a+1}{a^2-1} = a \][/tex]
### Part (e)
Simplify the expression:
[tex]\[ \frac{x}{x + y} - \frac{xy}{x^2 + 2xy + y^2} \][/tex]
#### Step 1: Recognize [tex]\( x^2 + 2xy + y^2 \)[/tex] as a perfect square
[tex]\[ x^2 + 2xy + y^2 = (x + y)^2 \][/tex]
So the expression becomes:
[tex]\[ \frac{x}{x + y} - \frac{xy}{(x + y)^2} \][/tex]
#### Step 2: Combine the fractions with a common denominator
The common denominator is [tex]\((x + y)^2\)[/tex]:
[tex]\[ \frac{x (x + y) - xy}{(x + y)^2} \][/tex]
#### Step 3: Simplify the numerator
Expand the numerator:
[tex]\[ x(x + y) = x^2 + xy \][/tex]
Subtract [tex]\(xy\)[/tex]:
[tex]\[ x^2 + xy - xy = x^2 \][/tex]
Thus, we have:
[tex]\[ \frac{x^2}{(x + y)^2} \][/tex]
Hence,
[tex]\[ \frac{x}{x + y} - \frac{xy}{x^2 + 2xy + y^2} = \frac{x^2}{(x + y)^2} \][/tex]
### Part (f)
Simplify the expression:
[tex]\[ \frac{1}{a - x} + \frac{x}{a^2 - 2ax - x} \][/tex]
#### Step 1: Simplify [tex]\(\frac{x}{a^2 - 2ax - x}\)[/tex]
Factor [tex]\(a^2 - 2ax - x\)[/tex]:
[tex]\[ a^2 - 2ax - x = a(a - x) - x(a - x) = (a - x)(a - x - 1) \][/tex]
So the expression becomes:
[tex]\[ \frac{x}{(a - x)(a - 1)} \][/tex]
#### Step 2: Combine the fractions with a common denominator
The common denominator is [tex]\((a - x)(a - 1)\)[/tex]:
[tex]\[ \frac{1(a - 1) + x}{(a - x)(a - 1)} \][/tex]
#### Step 3: Simplify the numerator
Expand [tex]\(1 \cdot (a - 1)\)[/tex]:
[tex]\[ a - 1 \][/tex]
Add [tex]\(x\)[/tex]:
[tex]\[ a - 1 + x \][/tex]
Thus, we have:
[tex]\[ \frac{a - 1 + x}{(a - x)(a - 1)} \][/tex]
Hence,
[tex]\[ \frac{1}{a - x}+\frac{x}{a^2-2ax-x} = \frac{a + x - 1}{(a - x)(a - 1)} \][/tex]
You now have the simplified forms for each expression:
d) [tex]\(a\)[/tex]
e) [tex]\(\frac{x^2}{(x+y)^2}\)[/tex]
f) [tex]\(\frac{a + x - 1}{(a - x)(a - 1)}\)[/tex]
### Part (d)
Simplify the expression:
[tex]\[ \frac{a^2 + 1}{a + 1} + \frac{a^2 - a + 1}{a^2 - 1} \][/tex]
#### Step 1: Simplify [tex]\(\frac{a^2 - a + 1}{a^2 - 1}\)[/tex]
Notice that [tex]\(a^2 - 1\)[/tex] is a difference of squares:
[tex]\[ a^2 - 1 = (a - 1)(a + 1) \][/tex]
Therefore, we have:
[tex]\[ \frac{a^2 - a + 1}{(a - 1)(a + 1)} \][/tex]
#### Step 2: Combine the fractions with a common denominator
The common denominator for the two fractions is [tex]\((a + 1)(a - 1)\)[/tex].
[tex]\[ \frac{a^2 + 1}{a + 1} = \frac{(a^2 + 1)(a - 1)}{(a + 1)(a - 1)} \][/tex]
This gives us:
[tex]\[ \frac{(a^2 + 1)(a - 1) + (a^2 - a + 1)}{(a + 1)(a - 1)} \][/tex]
#### Step 3: Simplify the numerator
Expand [tex]\( (a^2 + 1)(a - 1) \)[/tex]:
[tex]\[ (a^2 + 1)(a - 1) = a^3 - a^2 + a - 1 \][/tex]
Add this result to [tex]\(a^2 - a + 1\)[/tex]:
[tex]\[ a^3 - a^2 + a - 1 + a^2 - a + 1 \][/tex]
Combine like terms:
[tex]\[ a^3 - a^2 + a^2 + a - a - 1 + 1 = a^3 \][/tex]
Thus, we have:
[tex]\[ \frac{a^3}{(a + 1)(a - 1)} = \frac{a^3}{a^2 - 1} \][/tex]
Since [tex]\(a^3 = a \cdot (a^2)\)[/tex], this fraction simplifies to:
[tex]\[ a \ \text{(as the terms cancel out)} \][/tex]
Hence,
[tex]\[ \frac{a^2+1}{a+1}+\frac{a^2-a+1}{a^2-1} = a \][/tex]
### Part (e)
Simplify the expression:
[tex]\[ \frac{x}{x + y} - \frac{xy}{x^2 + 2xy + y^2} \][/tex]
#### Step 1: Recognize [tex]\( x^2 + 2xy + y^2 \)[/tex] as a perfect square
[tex]\[ x^2 + 2xy + y^2 = (x + y)^2 \][/tex]
So the expression becomes:
[tex]\[ \frac{x}{x + y} - \frac{xy}{(x + y)^2} \][/tex]
#### Step 2: Combine the fractions with a common denominator
The common denominator is [tex]\((x + y)^2\)[/tex]:
[tex]\[ \frac{x (x + y) - xy}{(x + y)^2} \][/tex]
#### Step 3: Simplify the numerator
Expand the numerator:
[tex]\[ x(x + y) = x^2 + xy \][/tex]
Subtract [tex]\(xy\)[/tex]:
[tex]\[ x^2 + xy - xy = x^2 \][/tex]
Thus, we have:
[tex]\[ \frac{x^2}{(x + y)^2} \][/tex]
Hence,
[tex]\[ \frac{x}{x + y} - \frac{xy}{x^2 + 2xy + y^2} = \frac{x^2}{(x + y)^2} \][/tex]
### Part (f)
Simplify the expression:
[tex]\[ \frac{1}{a - x} + \frac{x}{a^2 - 2ax - x} \][/tex]
#### Step 1: Simplify [tex]\(\frac{x}{a^2 - 2ax - x}\)[/tex]
Factor [tex]\(a^2 - 2ax - x\)[/tex]:
[tex]\[ a^2 - 2ax - x = a(a - x) - x(a - x) = (a - x)(a - x - 1) \][/tex]
So the expression becomes:
[tex]\[ \frac{x}{(a - x)(a - 1)} \][/tex]
#### Step 2: Combine the fractions with a common denominator
The common denominator is [tex]\((a - x)(a - 1)\)[/tex]:
[tex]\[ \frac{1(a - 1) + x}{(a - x)(a - 1)} \][/tex]
#### Step 3: Simplify the numerator
Expand [tex]\(1 \cdot (a - 1)\)[/tex]:
[tex]\[ a - 1 \][/tex]
Add [tex]\(x\)[/tex]:
[tex]\[ a - 1 + x \][/tex]
Thus, we have:
[tex]\[ \frac{a - 1 + x}{(a - x)(a - 1)} \][/tex]
Hence,
[tex]\[ \frac{1}{a - x}+\frac{x}{a^2-2ax-x} = \frac{a + x - 1}{(a - x)(a - 1)} \][/tex]
You now have the simplified forms for each expression:
d) [tex]\(a\)[/tex]
e) [tex]\(\frac{x^2}{(x+y)^2}\)[/tex]
f) [tex]\(\frac{a + x - 1}{(a - x)(a - 1)}\)[/tex]
