1). (2,3х²y + 1,1xy + 6y²) - (4,1xy - 1,2x²y + 6y²) =
= 2,3х²y + 1,1xy + 6y² - 4,1xy + 1,2x²y - 6y² = 3,5x²y - 3xy = xy(3,5x - 3)
при х = 2, у = 3:
xy(3,5x - 3) = 2 · 3 · (3,5 · 2 - 3) = 6 · 4 = 24
при х = -1, у = 4:
xy(3,5x - 3) = -1 · 4 · (3,5 · (-1) - 3) = -4 · (-6,5) = 26
2). А. 2 - (1,2х - 14,4) = 10 + 2х
2 + 14,4 - 10 = 1,2х + 2х
3,2х = 6,4
х = 2
Б). 5,6 - 1,2у + (3,4у - 0,2) = 5,4у + 11,8
- 1,2у + 3,4у - 5,4у = 11,8 - 5,6 + 0,2
-3,2у = 6,4
у = -2
x1=πn,n∈z
3π<πn<4π
3<n<4
нет решения
6cos²x-11cosx+4=0
cosx=a
6a²-11a+4=0
D=121-96=25
a1=(11-5)/12=1/2⇒cosx=1/2⇒x=11π/6+2πk,k∈z
3π<11π/6+2πk<4π
18<11+12k<24
7<12k<13
7/12<k<13/12
k=1⇒x=11π/6+2π=23π/6
a2=(11+5)/12=4/3⇒cosx=4/3>1 нетрешения
2)2сos²x+10sin2xcos2x+4sin²x+4cos²x=0/cos²x
4tg²x+10tgx+6=0
tgx=a
2a²+5a+3=0
D=25-24=1
a1=(-5-1)/4=-1,5⇒tgx=-1,5⇒x=-arctg1,5+πn
x=2π-arctg1,5
a2=(-5+1)/4=-1⇒tgx=-1⇒x=-π/4+πk,k∈z
x=3π/4
3)3cos²x+5sinxcosx+2cos²x=0
5cosx*(cosx+sinx)=0
cosx=0⇒x=π/2+πn,n∈z
x=5π/2
cosx+sinx=0/cosx
tgx+1=0
tgx=-1⇒x=-π/4+πm,m∈z
x=7π/4