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$1)\; \; \frac{1}{4k-5}-\frac{2}{5+4k}-\frac{2k-3}{25-16k^2}=\frac{5+4k-8k+10-2k+3}{(5-4k)(5+4k)}=\frac{-6k+18}{(5-4k)(5+4k)}=\\\\=\frac{-6(k-3)}{(5-4k)(5+4k)}\\\\2)\; \; \frac{5}{3a-b}+\frac{4}{3a+b}-\frac{2a-3b}{(3a-b)^2}=\frac{5(9a^2-b^2)+4(9a^2-6ab+b^2)-(2a-3b)(3a+b)}{(3a-b)^2(3a+b)}=\\\\=\frac{45a^2-5b^2+36a^2-24ab+4b^2-(6a^2+2ab-9ab-3b^2)}{(3a-b)^2(3a+b)}=\frac{75a^2+2b^2-17ab}{(3a-b)^2(3a+b)}$

$3)\; \; \frac{4-1,25a}{a^2-9}-\frac{3}{4a-12}-\frac{1}{8a+24}=\frac{4-1,25a}{(a-3)(a+3)}-\frac{3}{4(a-3)}-\frac{1}{8(a+3)}=\\\\=\frac{8(4-1,25a)-6(a+3)-(a-3)}{8(a-3)(a+3)}=\frac{32-10a-6a-18-a+3}{8(a-3)(a+3)}=\frac{17-17a}{8(a-3)(a+3)}=\\\\=\frac{17(1-a)}{8(a-3)(a+3)}$

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vologinavaleri 2) 5/3a-b + 4/3a+b – 2a+3b/(3a-b)²=

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Verified answer

1)1/4k-5 – 2/5+4k – 2k-3/25-16k²=
–1/5+4k – 2/5+4k – 2k-3/(5-4k)(5+4k)=
(– 3/5+4k)×5-4k — 2k-3/(5-4k)(5+4k)=
—3(5-4k)—2k-3 =-15+20k-2k-3=18k-18
Ответ: 18k-18.

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