Решение 3 и 4 на фотографии
1 5/6+2 1/12 *(1.15-1.23/0.6)=11/6+25/12*(-0.9)=11/6-25*9/120=11/6-5*9/24=11/6-5*3/6=-4/6=-2/3
12b/(b+4)(b^2-4b+16)-b+4/b^2-4b+16=12b-(b+4)^2/(b+4)(b^2-4b+16)=(12b-b^2-8b-16)/(b+4)(b^2-4b+16)=(-b^2+4b-16)/(b+4)(b^2-4b+16)= -(b^2-4b+16)/
/(b+4)(b^2-4b+16)= -1/ b+4
1)Х^4-28х^2+100=0
t=x^2
t1+t2=29 |t1=4
t1×t2=100 |t2=25
Ответ:t1=4 t2=25
2)t1=5 t2=2
1)<span>cos(a+b)+2sinasinb/2sinacosb-sin(a+b)=
(</span>cosacosb-sinasinb+2sinasinb)/(2sinacosb-sinacosb-cosasinb)=
=(cosacosb+sinasinb)/(sinacosb-cosasinb)=cos(a-b)/sin(a-b)=ctg(a-b)