Step 1

We can simplify the algebraic expressions involving fractions by taking LCM (Least common multiple ) of the denominators . We have to convert all the denominators to the value of the LCM by multiplying both numerator and denominator in each fraction by using suitable constant.

We can also factor out the common terms from both numerator and denominator of the expression to get the simple fraction form the algebraic expression . We can cancel the factors which are in numerator and denominator of equal kind.

Step 2

Consider the algebraic expression \(\displaystyle{\left[{\frac{{{x}}}{{{x}+{1}}}}\right]}-{\left[{\frac{{{2}{x}-{2}}}{{{x}-{1}}}}\right]}\),

We have to take the LCM of the denominator and we need to convert both the denominators of the fractions to the LCM value by multiplying both numerator and denominator in the expression by suitable terms.

We have the LCM of the denominators \(\displaystyle{\left({x}+{1}\right)}{\left({x}-{1}\right)}\) is \(\displaystyle{\left({x}+{1}\right)}{\left({x}-{1}\right)}\) .

To convert both denominators to LCM value , multiply both numerator and denominator of the first fraction by \(\displaystyle{\left({x}-{1}\right)}\) and the second fraction by \(\displaystyle{\left({x}+{1}\right)}\).

Thus we get,

\(\displaystyle{\left[{\frac{{{x}}}{{{x}+{1}}}}\right]}-{\left[{\frac{{{2}{x}-{2}}}{{{x}-{1}}}}\right]}\)

\(=\frac{x}{x+1} \times \frac{(x-1)}{(x-1)}-\frac{2x-3}{x-1} \times \frac{(x+1)}{(x+1)}\)

\(\displaystyle={\frac{{{x}{\left({x}-{1}\right)}-{\left({2}{x}-{3}\right)}{\left({x}+{1}\right)}}}{{{\left({x}+{1}\right)}{\left({x}-{1}\right)}}}}\)

\(\displaystyle={\frac{{{x}^{{{2}}}-{x}-{2}{x}{\left({x}+{1}\right)}+{3}{\left({x}+{1}\right)}}}{{{x}^{{{2}}}-{x}+{x}-{1}}}}\)

\(\displaystyle={\frac{{{x}^{{{2}}}-{x}-{2}{x}^{{{2}}}-{2}{x}+{3}{x}+{3}}}{{{x}^{{{2}}}-{1}}}}\) (simplify by adding the like terms)

\(\displaystyle={\frac{{-{x}^{{{2}}}+{3}}}{{{x}^{{{2}}}-{1}}}}\)

\(\displaystyle={\frac{{-{\left({x}^{{{2}}}-{3}\right)}}}{{{x}^{{{2}}}-{1}}}}\)

\(\displaystyle=-{\frac{{{\left({x}^{{{2}}}-{3}\right)}}}{{{x}^{{{2}}}-{1}}}}\)

Hence we have the required simple fraction form.