Question

допоможіііііііііііть​

Answer 1

$\frac{3}{t^{2} }:\frac{6}{t}= \frac{3}{t^{2} }*\frac{t}{6}=\frac{1}{2t} \\\\2)18*\frac{y}{2x^{2}} =\frac{2*9*y}{2x^{2}}=\frac{9y}{x^{2} }\\\\(-\frac{2a}{9b^{3}})^{2}=\frac{4a^{2} }{81b^{6}}\\\\\frac{x-y}{a+3}:\frac{x^{2}-2xy+y^{2}}{a^{2}+3a}=\frac{x-y}{a+3}*\frac{a(a+3)}{(x-y)^{2}}=\frac{a}{x-y}$

$3)\frac{x^{2} +4}{x^{2}-2x+4 }:\frac{16-x^{2}}{8+x^{3}}=\frac{x^{2} +4}{x^{2}-2x+4}*\frac{(2+x)(x^{2}-2x+4)}{(4-x^{2})(4+x^{2})}=\frac{2+x}{4-x^{2}}=\frac{2+x}{(2+x)(2-x)}=\frac{1}{2-x} \\\\4)\frac{a^{2}+ab-3a-3b }{a^{2}+ab+3a+3b}*\frac{4a+12}{8a-24}=\frac{(a^{2}+ab)-3(a+b) }{(a^{2}+ab)+3(a+b)}*\frac{4(a+3)}{8(a-3)}=\frac{a(a+b)-3(a+b)}{a(a+b)+3(a+b)} *\frac{(a+3)}{2(a-3)}=\\\\=\frac{(a+b)(a-3)}{(a+b)(a+3)}*\frac{a+3}{2(a-3)}=\frac{1}{2}$