Ответ:
[tex]-\frac{8}{2a-3}[/tex]
Объяснение:
[tex]\frac{2a}{2a+3}+\frac{5}{3-2a}-\frac{4a^2+9}{4a^2-9}=\frac{2a}{2a+3}-\frac{5}{2a-3}-\frac{4a^2+9}{(2a-3)(2a)+3}=\\\\\frac{2a(2a-3)}{(2a-3)(2a+3)}-\frac{5(2a+3)}{(2a-3)(2+3)}-\frac{4a^2+9}{(2a-3)(2a+3)}=\\\\\frac{2a(2a-3)-5(2a+3)-(4a^2+9)}{(2a-3)(2a)+3}=\frac{4a^2-6a-10a-15-4a^2-9}{(2a-3)(2a+3)}=\\\\\frac{-16a-24}{(2a-3)(2a+3)}=\frac{-8(2a+3)}{(2a-3)(2a+3)}=-\frac{8}{2a-3}[/tex]
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Answers & Comments
Ответ:
[tex]-\frac{8}{2a-3}[/tex]
Объяснение:
[tex]\frac{2a}{2a+3}+\frac{5}{3-2a}-\frac{4a^2+9}{4a^2-9}=\frac{2a}{2a+3}-\frac{5}{2a-3}-\frac{4a^2+9}{(2a-3)(2a)+3}=\\\\\frac{2a(2a-3)}{(2a-3)(2a+3)}-\frac{5(2a+3)}{(2a-3)(2+3)}-\frac{4a^2+9}{(2a-3)(2a+3)}=\\\\\frac{2a(2a-3)-5(2a+3)-(4a^2+9)}{(2a-3)(2a)+3}=\frac{4a^2-6a-10a-15-4a^2-9}{(2a-3)(2a+3)}=\\\\\frac{-16a-24}{(2a-3)(2a+3)}=\frac{-8(2a+3)}{(2a-3)(2a+3)}=-\frac{8}{2a-3}[/tex]