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

chicagochica5
 2 years ago
Best ResponseYou've already chosen the best response.0The first part would be (x 2) 2x? + 1

chicagochica5
 2 years ago
Best ResponseYou've already chosen the best response.0@theredhead1617 do you know how to solve the problem

miteshchvm
 2 years ago
Best 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?

chicagochica5
 2 years ago
Best ResponseYou've already chosen the best response.0No @miteshchvm I dont understand how to. I wish to know where to begin but i do not

miteshchvm
 2 years ago
Best ResponseYou've already chosen the best response.1can you factorise x^2 + 2x + 1?

miteshchvm
 2 years ago
Best ResponseYou've already chosen the best response.1x^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?

chicagochica5
 2 years ago
Best ResponseYou've already chosen the best response.0(x + 1) to second power?

miteshchvm
 2 years ago
Best 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,

chicagochica5
 2 years ago
Best ResponseYou've already chosen the best response.0just a little bit, its hard

miteshchvm
 2 years ago
Best ResponseYou've already chosen the best response.1now refer to my first comment

chicagochica5
 2 years ago
Best ResponseYou've already chosen the best response.0on the previous question?

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

chicagochica5
 2 years ago
Best ResponseYou've already chosen the best response.0im getting a better understand of it

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

chicagochica5
 2 years ago
Best ResponseYou've already chosen the best response.0thanks for everything @miteshchvm
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