1) (7x+x^2) /(12x-1)≤0
(x(7 + x)) /(12x-1)≤0
y=x(7 +x)*(12x-1); 12x-1≠0; x≠1/12
x(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...
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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 +πn
x⊂[0;π/2] 1-я четверть 2-ая четверть
x=π/12+πn; n-celoe
0≤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/12
3)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 наименьшее
f(-1)=1+1/2+1=2,5; наибольшее
f(1)=1-1/2+1=1,5
5) √(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=0
x=-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<0<br>sin(x/2-π/4)<-√2 /2<br>t=x/2-π/4; sint<-√2/2<br> 5π/4+2πn 5π/4+ 2πn 6π/4+2πn 1,5π+2πn < x/2<2π+2πn | *2<br> 3π+4πn ------------------------------------