1) (7x+x^2) /(12x-1)≤0 (x(7 + x)) /(12x-1)≤0y=x(7 +x)*(12x-1); 12x-1≠0; x≠1/12x(7 +x)(12x-1)=0; x=0 ili 7+x=0 i x≠1/12 (отметим кругляшком!) - х=-7 + - +------------(-7)-------------0----------1/12-------------x f(1)=1(7+1)(12-1)>0...//////////////// //////////////////x⊂(-∞;-7] ∪[0;1/12)2) 2sin(2x) -1=0; [0;π/2] 2sin(2x)=1; sin2x=1/2; 2x=π/6+2πn; 2x=5π/6+2πn; n-celoe x=π/12+πn; x=5π/12 +πnx⊂[0;π/2] 1-я четверть 2-ая четвертьx=π/12+πn; n-celoe0≤π/12+πn≤π/2 -------------------π/2≤πn≤π/2-π/12 |:π -1/2≤ n ≤ 5/12; n-celoe n=0; x=π/12 Ответ. π/123)y=x^4-x/2+1; [-1;1] наиб(наим)? D(y)=(-∞;+∞)y'=4x^3 -1/2; 4x^3- 1/2=0; x^3 = 1/8 ; x=1/2; 1/2⊂[-1;1]f(1/2)=(1/2)^4 -1/4+1=1/16-4/16+1=1-3/16=13/16 наименьшееf(-1)=1+1/2+1=2,5; наибольшееf(1)=1-1/2+1=1,55) √(9-x^2) *cos2x =0 - + - 9-x^2≥0; (3-x)(3+x)≥0 ----------------(-3)------------3------------->x ///////////////////// x⊂[-3;3]9-x^2=0 ili cos2x=0x=-3;3 2x=π/2+πn; x=π/4+πn; n-celoe x=π/4; x=3/4 x=π/4+π; x=3(3/4) ⊄[-3;3]x=-3; 3; π/4 всего три корня!6) 1/сos20 - 4sin50=1/cos20 - 4* sin(90-40)=1/cos20 -4*cos40==(1-4cos40 *cos20) /cos20=(1-4*((cos60)+cos20)/2) ) /cos20==(1-2*1/2 -2cos20) / cos20=-2cos20 / cos20=-2;7) y=sin(x/2-π/4) +√2 /2; ниже оси х sin(x/2 -π/4) +√2 /2<0sin(x/2-π/4)<-√2 /2t=x/2-π/4; sint<-√2/2 5π/4+2πn<t< 7π/4+2πn; n-celoe 5π/4+ 2πn<x/2 -π/4< 7π/4 +2πn |+π/4 6π/4+2πn<x/2 <8π/4 +2πn 1,5π+2πn < x/2<2π+2πn | *2 3π+4πn<x<4π+4πn ili π+4πn<2π+4πn; n-celoe ------------------------------------