27. Выполните действия: 1) (x + 4)/(2(x − 3)) − (x + 1)/(x − 3); 2) (a + 3)/(3a − 3) + (2 − a)/(5a − 5); 3) (x + 5)/(x − 5) − (x − 1)/(x + 5); 4) 4b/(3b − 21) + 3b/(14 − 2b); 5) 3p/(3p + 2q) − 9p²/(9p² + 12pq + 4q²); 6) 4/(c² − 36) − 2/(c² − 6c).

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Accepted Answer

При \(x\ne3\): \(\frac{x+4}{2(x-3)}-\frac{x+1}{x-3}=\frac{x+4-2(x+1)}{2(x-3)}=\frac{2-x}{2(x-3)}\).
При \(a\ne1\): \(\frac{a+3}{3a-3}+\frac{2-a}{5a-5}=\frac{5(a+3)+3(2-a)}{15(a-1)}=\frac{2a+21}{15(a-1)}\).
При \(x\ne\pm5\): \(\frac{x+5}{x-5}-\frac{x-1}{x+5}=\frac{(x+5)^2-(x-1)(x-5)}{x^2-25}=\frac{16x+20}{x^2-25}=\frac{4(4x+5)}{x^2-25}\).
При \(b\ne7\): \(\frac{4b}{3(b-7)}+\frac{3b}{-2(b-7)}=\frac{8b-9b}{6(b-7)}=-\frac{b}{6(b-7)}\).
Так как \(9p^2+12pq+4q^2=(3p+2q)^2\), при \(3p+2q\ne0\): \(\frac{3p}{3p+2q}-\frac{9p^2}{(3p+2q)^2}=\frac{3p(3p+2q)-9p^2}{(3p+2q)^2}=\frac{6pq}{(3p+2q)^2}\).
Разложим знаменатели на множители. При \(c\ne0,\pm6\): \(\frac{4}{(c-6)(c+6)}-\frac{2}{c(c-6)}=\frac{4c-2(c+6)}{c(c-6)(c+6)}=\frac{2(c-6)}{c(c-6)(c+6)}=\frac{2}{c(c+6)}\).

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