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agressia666
@agressia666
August 2021
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ТРИГОНОМЕТРИЯ. 40 БАЛЛОВ!
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sedinalana
Verified answer
2.1
f(x)=x²/(x-1)
f`(x)=(2x*(x-1)-1*x²)/(x-1)²=(2x²-2x-x²)/(x-1)²=(x²-2x)/(x-1)²
2.2
4
2.3
sin²x+4sinxcosx+3cos²x=0/cos²x
tg²x+4tgx+3=0
tgx=a
a²+4a+3=0
a1+a2=-4 U a1*a2=3
a1=-3⇒tgx=-3⇒x=-arctg3+πn,n∈z
a2=-1⇒tgx=-1⇒x=-π/4+πk,k∈z
2.4
S=πR²+πRL
S=200π,L=17
π(R²+17R)=200π
R²+17R-200=0
R1+R2=-17 U R1*r2=-200
R1=-25 не удов усл
R2=8
V=1/3*πR²*h
h=√(L²-R²)=√(289-64)=√225=15
V=1/3*π*64*15=320π
3,1
(4sin²acos²a-4sin²a)/(4sin²acos²a+4(sin²a-1))=
=4sin²a(cos²a-1)/(4sin²acos²a-4cos²a)=
=4sin²a*(-sin²a)/[4cos²a(sin²a-1)]=-4sin^4a/(-4cos^4a)=tg^4a
3,2
f(x)=(x+3)/(x-2)
f(3)=6
f`(x)=(1*(x-2)-1*(x+3))/(x-2)²=(x-2-x-3)/(x-2)²=-5/(x-2)²
f`(3)=-5
Y=6-5*(x-3)=6-5x+15=21-5x
x=0⇒y=21
y=0⇒21-5x=0⇒5x=21⇒x=4,2
Получили треугольник с вершинами (0;0);(0;21);(4,2;0)-прямоугольный
Длины катетов 21 и 4,2
S=1/2*21*4,2=44,1
2 votes
Thanks 1
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Answers & Comments
Verified answer
2.1f(x)=x²/(x-1)
f`(x)=(2x*(x-1)-1*x²)/(x-1)²=(2x²-2x-x²)/(x-1)²=(x²-2x)/(x-1)²
2.2
4
2.3
sin²x+4sinxcosx+3cos²x=0/cos²x
tg²x+4tgx+3=0
tgx=a
a²+4a+3=0
a1+a2=-4 U a1*a2=3
a1=-3⇒tgx=-3⇒x=-arctg3+πn,n∈z
a2=-1⇒tgx=-1⇒x=-π/4+πk,k∈z
2.4
S=πR²+πRL
S=200π,L=17
π(R²+17R)=200π
R²+17R-200=0
R1+R2=-17 U R1*r2=-200
R1=-25 не удов усл
R2=8
V=1/3*πR²*h
h=√(L²-R²)=√(289-64)=√225=15
V=1/3*π*64*15=320π
3,1
(4sin²acos²a-4sin²a)/(4sin²acos²a+4(sin²a-1))=
=4sin²a(cos²a-1)/(4sin²acos²a-4cos²a)=
=4sin²a*(-sin²a)/[4cos²a(sin²a-1)]=-4sin^4a/(-4cos^4a)=tg^4a
3,2
f(x)=(x+3)/(x-2)
f(3)=6
f`(x)=(1*(x-2)-1*(x+3))/(x-2)²=(x-2-x-3)/(x-2)²=-5/(x-2)²
f`(3)=-5
Y=6-5*(x-3)=6-5x+15=21-5x
x=0⇒y=21
y=0⇒21-5x=0⇒5x=21⇒x=4,2
Получили треугольник с вершинами (0;0);(0;21);(4,2;0)-прямоугольный
Длины катетов 21 и 4,2
S=1/2*21*4,2=44,1