Question

a rog punctele b si e dar pe lung daca bi stiti nu rezolvati

Answer 1

    
[tex]\displaystyle b) \\ \\ \frac{ \sqrt{3} x^2 }{2x-2 \sqrt{2} } \cdot \frac{ \sqrt{2} x-2}{3x^3}= \frac{ \sqrt{3} x^2 }{2(x- \sqrt{2}) } \cdot \frac{ \sqrt{2} (x-\sqrt{2})}{3x^3}= \\ \\ = \frac{ \sqrt{3} x^2 \cdot \sqrt{2} (x-\sqrt{2})}{3x^3 \cdot 2(x- \sqrt{2}) }= \frac{ \sqrt{3} \cdot \sqrt{2} }{3x \cdot 2}= \boxed{\frac{ \sqrt{6} }{6x }} [/tex]


[tex]\displaystyle e) \\ \\ \frac{(y-5)^2}{3y+18} : \frac{3y^2-75}{y^2-36}= \frac{(y-5)^2}{3y+18} \cdot \frac{y^2-36}{3y^2-75}= \\ \\ =\frac{(y-5)^2}{3(y+6)} \cdot \frac{y^2-36}{3(y^2-25)}= \\ \\ =\frac{(y-5)^2}{3(y+6)} \cdot \frac{(y-6)(y+6)}{3(y-5)(y+5)}= \\ \\ =\frac{(y-5)^2 \cdot(y-6)(y+6) }{ 3(y-5)(y+5) \cdot 3(y+6)} = \\ \\ =\frac{(y-5)\cdot(y-6) }{ 3(y+5) \cdot 3} =\boxed{\frac{(y-5)(y-6) }{ 9(y+5) } } [/tex]