Найти значение выражения. если tg a=-sqrt5 (5cos2a+3)/(3-8 cos^2 a) c обьяснением!

1)3 1/9 : 2 1/3 -2 5/6; а) 3 1/9: 2 1/3= 28/9 : 7/3=28/9*3/7=84/63=4/3 б) 4/3- 2 5/6=-2+4/3-5/6=-2+8/6-5/6=-2+3/6=-9/6=-3/2=-1,5 ответ: -1,5 2)10 16/17: 8 5/11+ 1 2/3 а) 10 16/17: 8 5/11=186/17: 93/11= 186/17*11/93= 2046/1581= 22/17 б) 22/17+ 1 2/3= 1+22/17+2/3= 1+66/51+34/51= 1 100/51= 2 49/51 ответ: 2 49/51 3) 1 5/7-4 3/13: 1 19/26 а) 4 3/13: 1 19/26= 55/13: 45/26=55/13*26/45= 1430/585= 22/9 б) 1 5/7-22/9=1+45/63-154/63= 1-109/63= -46/63 ответ: -46/63 4)47/48: 3 13/27-13/16 а)47/48: 3 13/27=47/48: 94/27= 47/48*27/94=1269/4512=9/32 б)9/32-13/16=9/32-26/32=-17/32 ответ: -17/32

X^2+2x-3> =0⇒(x+3)(x-1)≥0⇒x≤-3 u x≥1 x^2+16x-3> =0⇒(x+8+√67)(x+8-√76)≥0⇒x≤-8-√67 u x≥-8+√67 x+3< =1⇒x≤-2 x²+2x-3=0 x1+x2=-2 u x1*x2=-3⇒x1=-3 u x2=1 x²+16x-3=0 d=256+12=268 x1=(-16-2√67)/2=-8-√67 u x2=-8+√67 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ -8-√---8+√ \\\\\\\\\\\\\\\                                                             /////////////////////////////////////   /////////////////////////////////////                                                               \\\\\\\\\\\\\\\\\   x∈(-∞; -8-√67]