Y=4x-12/3
3x+5*1/3x-16=-24
8*1/3x=-8
y=4x-12/3
x=-24/25
y=32/25-4
x=-0.96
y=-2.72
cosx+cos9x+cos5x=0
2cos((x+9x)/2)*cos((x-9x)/2)+cos5x=0
2cos5xcos(-4x)+cos5x=0
2cos5xcos4x+cos5x=0
cos5x(2cos4x+1)=0
1) cos5x=0
5x=(pi/2) + pi*k
x=(pi/10)+ (pi*k)/5
2) 2cos4x+1=0
2cos4x=-1
cos4x=-1/2
4x=+- arccos(-1/2)+2pi*k
4x=+- (2pi/3) + 2pi*k
x= +- (pi/6) + (pi*k)/2
Ответ: x= (pi/10)+(pi*k)/5 ; x=(pi/6)+(pi*k)/2 ; x= - (pi/6)+(pi*k)/2, k принадлежит Z
(tg92-tg32)/(1+tg92*tg32)=tg(92-32)=yg60=√3
<span>(х-2)(12-х)=9
12x-x^2-24+2x=9
14x-x^2-24-9=0
x^2-14x+33=0
D=(-14)^2-4*1*33==196-132=64=8^2
x1=(14+8)/2=11
x2=(14-8)/2=3
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</span><span>(х+1)^2=3(х+7)
x^2+2x+1=3x+21
x^2+2x+1-3x-21=0
x^2-x-20=0
D=(-1)^2-4*1*(-20)=1+80=81=9^2
x1=(1+9)/2=5
x2=(1-9)/2=-4
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