Solution:

\(m^2-2mn+n^2-9r^2.\)

\(=m^2-mn-mn+n^2-3^2r^2\)

\(=m(m-n)-n(m-n)-(3r)^2\)

\(=(m-n)(m-n)(3r)^2\)

\(=(m-n)^2-(3r)^2\)

\(=[(m-n)-3r][(m-n)+3r]\)

By: Abdul Karim

Solution:

\(9a^2-4(a-b)^2.\)

\(=3^2a^2-2^2(a-b)^2\)

\(=(3a)^2-[2(a-b)]^2\)

\(=[3a-2(a-b)][3a+2(a-b)]\)

By: Abdul Karim

Solution:

\(\frac{m+1}{m-1}-\frac{m-1}{m+1}+\frac{4}{m^2-1}\)

\(=\frac{m+1}{m-1}-\frac{m-1}{m+1}+\frac{4}{(m+1)(m-1)}\)

\(=\frac{(m+1)(m+1)-[(m-1)(m-1)]+4}{(m+1)(m-1)}\)

\(=\frac{m^2+2m+1-[m^2-2m+1]+4}{(m+1)(m-1)}\)

\(=\frac{m^2+2m+1-m^2+2m-1+4}{(m+1)(m-1)}\)

\(=\frac{m^2-m^2+2m+2m+1-1+4}{(m+1)(m-1)}\)

\(=\frac{4m+4}{(m+1)(m-1)}\)

\(=\frac{4(m+1)}{(m+1)(m-1)}\)

\(=\frac{4}{m-1}.\) Hint\((m+1)\) cancelled each other

By: Abdul Karim

Solution:

\(\frac{x+\frac{1}{x}+2}{x^2-1}.\)

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

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

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

\(=\frac{x^2+x+x+1}{x[(x+1)(x-1)]}\)

\(=\frac{x(x+1)+1(x+1)}{x(x+1)(x-1)}\)

\(=\frac{(x+1)(x+1)}{x(x+1)(x-1)}\)

\(=\frac{x+1}{x(x-1)}.\) Hint\((x+1)\) cancelled each other

By: Abdul Karim

Solution:

\(x^2+4x+3+mx+3m\)

\(=x^2+3x+x+3+mx+3m\)

\(=x(x+3)+1(x+3)+m(x+3)\)

\(=(x+1+m)(x+3).\)

By: Abdul Karim

Solution:

\(\frac{x^2+x-6}{x^2-3x+2}\times\frac{x^2-x}{x^2-9}\)

\(=\frac{x^2+3x-2x-6}{x^2-2x-x+2}\times\frac{x(x-1)}{x^2-3^2}\)

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

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

\(=\frac{x}{x-3}.\) Hint\((x+3)\), \((x-2)\) and \((x-1)\) have been cancelled

By: Abdul Karim