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miteshchvmBest ResponseYou've already chosen the best response.1
the x^2 + 2x + 1 is thw square of (x + 1)^2 as per the previous questions you can solve it
 one year ago

chicagochica5Best ResponseYou've already chosen the best response.0
The first part would be (x 2) 2x? + 1
 one year ago

chicagochica5Best ResponseYou've already chosen the best response.0
@theredhead1617 do you know how to solve the problem
 one year ago

miteshchvmBest ResponseYou've already chosen the best response.1
\[\frac{ \frac{ (x+1)(x+1) }{ x2 } }{ \frac{ (x1)(x+1) }{ (x2)(x+2) } }\] \[\frac{ (x+1)(x+1) }{ (x2) } \div { \frac{ (x1)(x+1) }{ (x2)(x+2) } }\\] \frac{ (x+1)(x+1) }{ (x2) } \times { \frac{ (x2)(x+2) }{ (x1)(x+1) } }\\]\] can you solv further?
 one year ago

chicagochica5Best ResponseYou've already chosen the best response.0
No @miteshchvm I dont understand how to. I wish to know where to begin but i do not
 one year ago

miteshchvmBest ResponseYou've already chosen the best response.1
can you factorise x^2 + 2x + 1?
 one year ago

miteshchvmBest ResponseYou've already chosen the best response.1
x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1) x^2  1 = (x +1)(x  1) x^2  4 = (x2)(x+2) you get it?
 one year ago

chicagochica5Best ResponseYou've already chosen the best response.0
(x + 1) to second power?
 one year ago

miteshchvmBest ResponseYou've already chosen the best response.1
\[x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1) \] \[x^2  1 = (x +1)(x  1) \] \[x^2  4 = (x2)(x+2)\] you get it? i wrote this instead of your question in order to simplify,
 one year ago

chicagochica5Best ResponseYou've already chosen the best response.0
just a little bit, its hard
 one year ago

miteshchvmBest ResponseYou've already chosen the best response.1
now refer to my first comment
 one year ago

chicagochica5Best ResponseYou've already chosen the best response.0
on the previous question?
 one year ago

miteshchvmBest ResponseYou've already chosen the best response.1
your question can be written as\[{ \frac{ (x^2 + 2x + 1) }{ x2 } } \div { \frac{ x^2  1 }{ x^2  4 } } \] now \[\frac{ (x+1)(x+1) }{ (x2) } \div { \frac{ (x1)(x+1) }{ (x2)(x+2) } }\\] \] \[\frac{ (x+1)(x+1) }{ (x2) } \times { \frac{ (x2)(x+2) }{ (x1)(x+1) } }\\] \]
 one year ago

chicagochica5Best ResponseYou've already chosen the best response.0
im getting a better understand of it
 one year ago

miteshchvmBest ResponseYou've already chosen the best response.1
on eliminating values you get \[\frac{ (x+1)(x+2) }{ (x1) }\]
 one year ago

chicagochica5Best ResponseYou've already chosen the best response.0
thanks for everything @miteshchvm
 one year ago
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