Answer 1

$1)\frac{t^{2}-2t+4 }{4t^{2}-1 }*\frac{2t^{2}+t }{t^{3}+8 }=\frac{t^{2}-2t+4}{(2t-1)(2t+1)}*\frac{t(2t+1)}{(t+2)(t^{2}-2t+4)}=\frac{t}{(2t-1)(t+2)}\\\\2)\frac{t}{(2t-1)(t+2)}-\frac{t+2}{2t^{2} -t}=\frac{t}{(2t-1)(t+2)}-\frac{t+2}{t(2t-1)}=\frac{t^{2}-t^{2}-4t-4}{t(2t-1)(t+2)}=\frac{-4t-4}{t(2t-1)(t+2)}$

$3)\frac{-4t-4}{t(2t-1)(t+2)}:\frac{7}{t^{2} +2t}=\frac{-4t-4}{t(2t-1)(t+2)}*\frac{t(t+2)}{7}=\frac{-4t-4}{7(2t-1)}=\frac{-4t-4}{14t-7}=\frac{4t+4}{7-14t}\\\\4)\frac{4t+4}{7-14t}-\frac{10t+1}{7-14t}=\frac{4t+4-10t-1}{7-14t}=\frac{3-6t}{7-14t}=\frac{3(1-2t)}{7(1-2t)}=\frac{3}{7} \\\\Otvet:\boxed{\frac{3}{7}}$

Answer 2

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