Question

№2 решить 1),2).заранее спс

Answer 1

$1)\frac{3a^{2}-12 }{3a^{2}+3}*(\frac{2a+1}{a-2}+\frac{2a-1}{a+2})=\frac{3(a^{2}-4)}{3(a^{2}+1)} *\frac{(2a+1)*(a+2)+(2a-1)*(a-2)}{(a-2)(a+2)}=\\\\=\frac{a^{2}-4 }{a^{2} +1} *\frac{2a^{2}+4a+a+2+2a^{2}-4a-a+2}{a^{2}-4}=\frac{4a^{2}+4 }{a^{2}+1} =\frac{4(a^{2}+1) }{a^{2}+1 }=4\\\\Otvet:\boxed{4}$

$2)(\frac{2x}{4x^{2}-12x+9 }-\frac{3}{4x^{2}-9 })*\frac{8x^{3}-18x }{4x^{2}+9 }=(\frac{2x}{(2x-3)^{2}}-\frac{3}{(2x-3)(2x+3)} )*\frac{2x(4x^{2}-9) }{4x^{2}+9}=\\\\=\frac{2x*(2x+3)-3*(2x-3)}{(2x-3)^{2}(2x+3)}*\frac{2x(4x^{2}-9) }{4x^{2}+9}=\frac{4x^{2}+6x-6x+9 }{(2x-3)(4x^{2}-9)}*\frac{2x(4x^{2}-9) }{4x^{2}+9}=\\\\=\frac{4x^{2}+9 }{(2x-3)(4x^{2}-9)}*\frac{2x(4x^{2}-9) }{4x^{2}+9}=\frac{2x}{2x-3}\\\\Otvet:\boxed{\frac{2x}{2x-3}}$