Предмет: Алгебра, автор: bisula738

a^2+3/a2-3a - 3a-2/a^2-3a + 4-3a/a^2-3a
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Ответы

Автор ответа: nafanya2014
1

\frac{a^2+3}{a^2-3a} -\frac{3a-2}{a^2-3a} +\frac{ 4-3a}{a^2-3a}=\frac{a^2+3-(3a-2)+(4-3a)}{a^2-3a} =\frac{a^2+3-3a+2+4-3a}{a^2-3a} =\frac{a^2-6a+9}{a^2-3a}=\\\\=\frac{(a-3)^2}{a(a-3)}=\frac{a-3}{a}

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