((m+n)/(m-1))+n ALL divided by ((m+n)/m-1))-1 please help :D
N nwelter New member Joined Aug 31, 2009 Messages 1 Aug 31, 2009 #1 ((m+n)/(m-1))+n ALL divided by ((m+n)/m-1))-1 please help
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Aug 31, 2009 #2 Multiply the top and bottom of the big fraction by (m−1)\displaystyle (m-1)(m−1).
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Aug 31, 2009 #3 Hello, nwelter! Simplify: m+nm−1+nm+nm−1−1\displaystyle \displaystyle{\text{Simplify: }\;\frac{\frac{m+n}{m-1}+n} {\frac{m+n}{m-1} -1} }Simplify: m−1m+n−1m−1m+n+n Click to expand... Multiply by m−1m−1 :(m−1)[m+nm−1+n](m−1)[m+nm−1−1] = (m+n)+n(m−1)(m+n)−(m−1) = m+n+mn−nm+n−m+1\displaystyle \displaystyle{\text{Multiply by }\frac{m-1}{m-1}\!:\quad\frac{(m-1)\left[\frac{m+n}{m-1} + n\right]} {(m-1)\left[\frac{m+n}{m-1} - 1\right]} \;=\;\frac{(m+n) + n(m-1)}{(m+n) - (m-1)} \;=\;\frac{m+ n + mn - n}{m + n - m + 1} }Multiply by m−1m−1:(m−1)[m−1m+n−1](m−1)[m−1m+n+n]=(m+n)−(m−1)(m+n)+n(m−1)=m+n−m+1m+n+mn−n . . . . . . . . . . = mn+mn+1 = m(n+1)n+1 = m\displaystyle = \;\frac{mn + m}{n + 1} \;=\;\frac{m(n+1)}{n+1} \;=\; m=n+1mn+m=n+1m(n+1)=m
Hello, nwelter! Simplify: m+nm−1+nm+nm−1−1\displaystyle \displaystyle{\text{Simplify: }\;\frac{\frac{m+n}{m-1}+n} {\frac{m+n}{m-1} -1} }Simplify: m−1m+n−1m−1m+n+n Click to expand... Multiply by m−1m−1 :(m−1)[m+nm−1+n](m−1)[m+nm−1−1] = (m+n)+n(m−1)(m+n)−(m−1) = m+n+mn−nm+n−m+1\displaystyle \displaystyle{\text{Multiply by }\frac{m-1}{m-1}\!:\quad\frac{(m-1)\left[\frac{m+n}{m-1} + n\right]} {(m-1)\left[\frac{m+n}{m-1} - 1\right]} \;=\;\frac{(m+n) + n(m-1)}{(m+n) - (m-1)} \;=\;\frac{m+ n + mn - n}{m + n - m + 1} }Multiply by m−1m−1:(m−1)[m−1m+n−1](m−1)[m−1m+n+n]=(m+n)−(m−1)(m+n)+n(m−1)=m+n−m+1m+n+mn−n . . . . . . . . . . = mn+mn+1 = m(n+1)n+1 = m\displaystyle = \;\frac{mn + m}{n + 1} \;=\;\frac{m(n+1)}{n+1} \;=\; m=n+1mn+m=n+1m(n+1)=m
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Sep 1, 2009 #4 f(x) = m+nm−1+nm+nm−1−1 = m+n+mn−nm−1m+n−m+1m−1\displaystyle f(x) \ = \ \frac{\frac{m+n}{m-1}+n}{\frac{m+n}{m-1}-1} \ = \ \frac{\frac{m+n+mn-n}{m-1}}{\frac{m+n-m+1}{m-1}}f(x) = m−1m+n−1m−1m+n+n = m−1m+n−m+1m−1m+n+mn−n = m(n+1)m−1∗m−1n+1 = m(n+1)n+1 = m\displaystyle = \ \frac{m(n+1)}{m-1}*\frac{m-1}{n+1} \ = \ \frac{m(n+1)}{n+1} \ = \ m= m−1m(n+1)∗n+1m−1 = n+1m(n+1) = m Restrictions: m≠ 1, n≠ −1.\displaystyle Restrictions: \ m\not= \ 1, \ n\not= \ -1.Restrictions: m= 1, n= −1.
f(x) = m+nm−1+nm+nm−1−1 = m+n+mn−nm−1m+n−m+1m−1\displaystyle f(x) \ = \ \frac{\frac{m+n}{m-1}+n}{\frac{m+n}{m-1}-1} \ = \ \frac{\frac{m+n+mn-n}{m-1}}{\frac{m+n-m+1}{m-1}}f(x) = m−1m+n−1m−1m+n+n = m−1m+n−m+1m−1m+n+mn−n = m(n+1)m−1∗m−1n+1 = m(n+1)n+1 = m\displaystyle = \ \frac{m(n+1)}{m-1}*\frac{m-1}{n+1} \ = \ \frac{m(n+1)}{n+1} \ = \ m= m−1m(n+1)∗n+1m−1 = n+1m(n+1) = m Restrictions: m≠ 1, n≠ −1.\displaystyle Restrictions: \ m\not= \ 1, \ n\not= \ -1.Restrictions: m= 1, n= −1.
