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valina [46]
4 years ago
8

Find the solutions of∅¬¬: 1/(x+a) - 1/(x+b) = 1/a x 1/b

Mathematics
1 answer:
VikaD [51]4 years ago
4 0
<span>1/(x+a) - 1/(x+b) = 1/a x 1/b

First, rewrite this as    </span><span>1/(x+a) - 1/(x+b) = (1/a)(1/b)  (do not use "x" to indicate                                                                                       mult.)

Then the LCD is (x+a)(x+b)(ab)

so the first term becomes (x+b)(ab), and
the second becomes (x+a)(ab), and 
the third becomes (x+a)(x+b).

Thus, we now have   (x+b)(ab) - (x+a)(ab) = (x+a)(x+b)(b) + (x+a)(x+b)(a).

Now, with the fractions gone, can you solve thios for x?</span>
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Answer:

1) {n}^{2}  - 81 \\ (n - 9)(n + 9)

2) \:  \: {a}^{2}  - 121 \\ (a - 11)(a + 11)

3) \:  \:  {n}^{2}  - 16 \\ (n - 4)(n + 4)

4)  \:  \: 9 {x}^{2}  - 144 \\  {3}^{2}  {x}^{2}  -  {12}^{2}  \\ (3x - 12)(3x + 12)

5) \: 2 {x}^{2}  - 9 \\ (2x - 3)(x + 3)

6) \:  \: 4 {w}^{2}  - 9 \\  {2}^{2}  {w}^{2}  -  {3}^{2}  \\ (2w + 3)(2w - 3)

7) \:  \: 4 {n}^{2}  - 1 \\  {2}^{2}  {n}^{2}  -  {1}^{2}  \\ (2n  + 1)(2n - 1)

8) \:  \: 1 - 16 {x}^{2}  \\  {1}^{2}  -  {4}^{2}  {x}^{2}  \\ (1 - 4x)(1  +  4x)

9) \:  \: {x}^{2}   - {y}^{2}  \\ (x - y)(x + y)

10) \:  \: 9 -  {c}^{2}  \\  {3}^{2}  -  {c}^{2}  \\ (3 - c)(3 + c)

11) \:  \:  {n}^{2} -  25 \\  {n}^{2}  -  {5}^{2}  \\ (n + 5)(n - 5)

13)49 - 4 {a}^{2}  \\  {7}^{2}  -  {2}^{2}  {a}^{2}  \\ (7 - 2a)(7 + 2a)

14) {a}^{2}  {b}^{2}  -  {c}^{4}  \\ (ab -  {c}^{2} )(ab +  {c}^{2} )

15) \:  \: 4  {x}^{2}  {y}^{2}  - 9 {z}^{2} \\  {2}^{2}   {x}^{2}  {y}^{2}  -  {3}^{2}  {z}^{2}  \\ (2xy - 3z)(2xy + 3z)

hope this helps

sorry, I don't know how solve the 12th one.

thanks.

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