Asked by anonymous
Please help with these 5 problems!
I have the solutions but I don't understand how to get them!
1. Perform the indicated operations and simplify:
[(2/x)-1]/(x^2 -4)
2. Subtract and simplify:
[12/(x^2 -4)] - [(3-x)/(x^2 + 2x)]
3. Perform the indicated operations and simplify:
[(x/y)-(y/x)]^-1
4. Divide and simplify:
[(a^2 b)/(a-b)]/[(a+b)/(a^2-b^2)]
5. Perform the indicated operations and simplify
[(x/x-3)-(2x/x^2-2X-3)]/[(2/x+1)-(1/x)
Solutions:
1. -1/(x^2 + 2x)
2. [x^2 + 7x + 6]/[x(x+z)(x-z)]
--typo in book?
3. xy/(x^2 - y^2)
4. a^2 b
5. x^2/(x-3)
I have the solutions but I don't understand how to get them!
1. Perform the indicated operations and simplify:
[(2/x)-1]/(x^2 -4)
2. Subtract and simplify:
[12/(x^2 -4)] - [(3-x)/(x^2 + 2x)]
3. Perform the indicated operations and simplify:
[(x/y)-(y/x)]^-1
4. Divide and simplify:
[(a^2 b)/(a-b)]/[(a+b)/(a^2-b^2)]
5. Perform the indicated operations and simplify
[(x/x-3)-(2x/x^2-2X-3)]/[(2/x+1)-(1/x)
Solutions:
1. -1/(x^2 + 2x)
2. [x^2 + 7x + 6]/[x(x+z)(x-z)]
--typo in book?
3. xy/(x^2 - y^2)
4. a^2 b
5. x^2/(x-3)
Answers
Answered by
drwls
1. Rewrite (2/x) -1 in the numerator as -(x-2)/x
Factor (x^2/4) in the denominator as (x-2)(x+2)
Then cancel out the (x-2) terms in numerator and denominator. You wil be left with -1/[x(x+2)]
3. Rewrite x/y and y/x with a common denominator, xy.
[(x^2- y^2)/xy]^-1 = xy/(x^2- y^2)
You try the others. They are all exercises in factoring and canceling tems
Factor (x^2/4) in the denominator as (x-2)(x+2)
Then cancel out the (x-2) terms in numerator and denominator. You wil be left with -1/[x(x+2)]
3. Rewrite x/y and y/x with a common denominator, xy.
[(x^2- y^2)/xy]^-1 = xy/(x^2- y^2)
You try the others. They are all exercises in factoring and canceling tems
Answered by
hhhhhhhhhhhh
8/9a^2 divide 4a^2-4a-24/a^2-6a+9 Perform the indicated operation
Answered by
Modest
(2/3-1/6)divide by(1/4+4/5) reduce to lowes terms
Answered by
Anonymous
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