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kondor19780726 [428]
3 years ago
15

Please anyone help me answer this question am giving the brainliest

Mathematics
1 answer:
DochEvi [55]3 years ago
6 0

Answer:

(-3/2x+1) + (3/x-1)

Step-by-step explanation:

Dont put the answer in parenthesis, I just did that to seprate it so it was easier to read. If you need it to be put in word form for it to be easier for you to understand, its "negative three divided by two x plus one, plus 3 divided by x minus 1"

First, rewrite the expression as 3x+6/2x^2+x-2x-1

Then, factor the expression, 3x+6/x*(2x+1)-(2x+1) (*=multiply)

Then, write a new fraction for each factor in the denominator, then write a new fraction, 3x+6/(2x+1)*(x-1) to (A)/2x+1 + B/x-1)

Then write terms in the numerators, for now we will use A and B.

Now we set the sum of the fractions equal to the original fraction.

3x+6/(2x+1)*(x-1) = A)/2x+1 + B/x-1).

Then reorder and group the terms, giving you 3x+6=(A+2B)x+(-A+B).

Now equate the coefficiants, {6= -A+B, 3= A+2B

Then substitute back, (A, B)=(-3, 3)

Now finally, just rewrite the fraction. (-3/2x+1) + (3/x-1)

Hope this helps! Sorry its kind of long.

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The change of base formula allows you to write a logarithm in terms of logarithms with another base. It follows this pattern,

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Assigning the base to be 7,

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I hope I was able to answer your question. Have a good day.
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