Ответ:(π/6+πk;π/3+πk),k∈z
Объяснение: sinx·cosx>√3/4;
2sinx·cosx>√3/2;
sin2x>√3/2;
π/3+2πk<2x<2π/3+2πk,k∈z;
π/6+πk<x<π/3+πk,k∈z.
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<span>-9у-12=-10у-6
</span><span>-9у+10y =12-6
</span>y =6
1)(m-4)(m+4)/m *1/[3(m+4)]=(m-4)/3m
2)(m-3)/2m-(m-4)/3m==(3m-9-2m+8)/6m=(m-1)/6m