A1. 10a^(1/2) +(√a -5)² =10√a +(a -10√a +25) =a+25.A2. Loq (10) (x-6) = -2 ;
Loq (10) (x-6) = Loq (10) ((10)^-2 ) ;x -6 = 10^ (-2) ;x =6+1/10² ;x =6,01.A3. Дано: sinα = - 0,5 ; 3π/2<α<2π.cosα -->? cosα = √(1-sin²α) =√(1 -0,5²) =√0,75= 0,5√3.cosα = √(1-sin²α) =√(1 -0,5²) =√(1-1/4) =(√3)/ 2 . * * * 3π/2<α<2π ⇒cosα > 0 * * *A4.√(12x -11) =x+2 ;{ x+2 ≥ 0 ;12x -11 =(x+2)² . { x≥ -2 ;x² -8x +15 =0. { x≥ -2 ;[ x=3; x =5.ответ : 3 ; 5 .A7. f(x) =x³ -3x +1 ; x∈[0 ;2]f'(x) =3x² -3x =3x(x-1);критические точки : f'(x) =0⇒ [ x=0 ;x=1. оба ∈[0 ;2]f(0) =1.f(1) =1³ -3*1 +1 = -1.f(2) =2³ -3*2 +1 =3.наименьшее значения функции f(x) =x³ -3x +1 в отрезке [0 ;2] : min y = -1.A10. AC₁ =√(A₁B₁²+A₁D₁² +BB₁²) =√(2²+8²+16²) =18.----------------------------------B1 .интеграл a=1 ,b=e (1/x*dx) =Lnx | a=1,b =e Lne -Ln1 =1-0 =1.B3. V =πR₁²H₁ =πR₂²H₂ ;R₁²H₁ =(7*R₁)²H₂ ;R₁²H₁ =49*R₁²H₂ ⇒H₂ =H₁/49 =98 см/49 =2 см .B4 2cos² x =√3sin(3π/2 +x) ;2cos²x = -√3scosx ;2cos²x+√3scosx =0 ;2cosx(cosx+√3/2) =0 ;[cosx =0 ; cosx+√3/2=0.[x = π/2+π*k ; x = ±5π/6 +2π*k , k∈Z.ответ : π/2+π*k ; ±5π/6 +2π*k , k∈Z.