Question

довести тотожность
$\frac{1}{6a - 4b} - \frac{1}{6a + 4b} - \frac{3a}{{4b}^{2} - {9a}^{2} } = \frac{1}{3a - 2b}$
$\frac{c + 2}{ {c}^{2} + 3c } - \frac{1}{3c + 9} - \frac{2}{3c} = 0$
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Answer 1

Ответ:

Объяснение:

1/(6a-4b) -1/(6a+4b) -(3a)/(4b²-9a²)=1/(3a-2b)

1/(2(3a-2b)) -1(2(3a+2b)) +(3a)/((3a-2b)(3a+2b))=1/(3a-2b)

(3a+2b-3a+2b+6a)/(2(3a-2b)(3a+2b))=1/(3a-2b)

(4b+6a)/(2(3a-2b)(3a+2b))=1/(3a-2b)

(2(3a+2b))/(2(3a-2b)(3a+2b))=1/(3a-2b)

1/(3a-2b)=1/(3a-2b)

(c+2)/(c²+3c) -1/(3c+9) -2/(3c)=0

(c+2)/(c(c+3)) -1/(3(c+3)) -2/(3c)=0

(3c+6-с-2с-6)/(3c(c+3))=0

0/(3c(c+3))=0

0=0