Question

Scrieți sub formă de raport algebric ireductibil:
a)$\frac{a^{2}-4a+4 }{b+b^{3} } : \frac{2-a}{b ^{2}+1 }$
b)$\frac{3a+2}{a- \sqrt{3} } : \frac{4+6a}{5a^{2} -15}$

Answer 1

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Answer 2

 [tex]a).\;\;...\;a^2-4a+4=(a-2)^2\;;\;b+b^3=b(b^2+1)\;;\\ \Rightarrow\;\frac{(a-2)^{\not2}}{b(b^2\not+1)}\cdot\frac{b^2\not+1}{-(a\not-2)}=\,-\;\frac{a-2}{b}\;;\\
b).\,4+6a=2(2+3a)\;;\;5a^2-15=5(a^2-3)=5(a+\sqrt3)(3-\sqrt3)\\ \Rightarrow\;...=\,\frac{3a\not+2}{a\not-\sqrt3}\cdot\frac{5(a+\sqrt3)(x\not-\sqrt3)}{2(2\not+3a)}=\frac{5(a+\sqrt3)}{2}\;;[/tex]