Объяснение:
В каком виде представлены выражения, в таком виде и будем решать:
(4ас^2)^3 •(0,5а^3 •с)^2=(2^2)^3 •(1/2)^2 •а^(3+3•2) •с^(2•3+2)=2^(2•3-2) •а^9 •с^8=2^4 •а^9 •с^8=16а^9 •с^8
(2/(3х^2 •у^3))^3 •(-9х^4)^2=8/3^3 •(-(3^2))^2 •х^(-2•3+4•2) •у^(-3•3)=8•3^(-3+2•2) •х^(-6+8) •у^(-9)=(8•3)/(х^2 •у^9)=24/(х^2 •у^9)
-(-х^2 •у^4)^4 •(6х^4 •у)^2=-36х^(2•4+4•2) •у^(4•4+2)=-36х^(8+8) •у^18=-36х^16 •у^18
(-10а^3 •b^2)^5 •(-0,2ab^2)^5=(-10)^5 •(-2/10)^5 •a^(3•5+5) •b^(2•5+2•5)=32•10^(5-5) •a^20 •b^(10+10)=32a^20 •b^20
Based on two different cases:
x
=
π
6
,
5
or
3
2
Look below for the explanation of these two cases.
Explanation:
Since,
cos
+
sin
1
we have:
−
So we can replace
in the equation
by
(
)
⇒
or,
0
using the quadratic formula:
b
±
√
4
a
c
for quadratic equation
⋅
8
9
Case I:
for the condition:
≤
to get positive value of
Case II:
to get negative value of
Answer link
Объяснение:
В каком виде представлены выражения, в таком виде и будем решать:
(4ас^2)^3 •(0,5а^3 •с)^2=(2^2)^3 •(1/2)^2 •а^(3+3•2) •с^(2•3+2)=2^(2•3-2) •а^9 •с^8=2^4 •а^9 •с^8=16а^9 •с^8
(2/(3х^2 •у^3))^3 •(-9х^4)^2=8/3^3 •(-(3^2))^2 •х^(-2•3+4•2) •у^(-3•3)=8•3^(-3+2•2) •х^(-6+8) •у^(-9)=(8•3)/(х^2 •у^9)=24/(х^2 •у^9)
-(-х^2 •у^4)^4 •(6х^4 •у)^2=-36х^(2•4+4•2) •у^(4•4+2)=-36х^(8+8) •у^18=-36х^16 •у^18
(-10а^3 •b^2)^5 •(-0,2ab^2)^5=(-10)^5 •(-2/10)^5 •a^(3•5+5) •b^(2•5+2•5)=32•10^(5-5) •a^20 •b^(10+10)=32a^20 •b^20
Based on two different cases:
x
=
π
6
,
5
π
6
or
3
π
2
Look below for the explanation of these two cases.
Explanation:
Since,
cos
x
+
sin
2
x
=
1
we have:
cos
2
x
=
1
−
sin
2
x
So we can replace
cos
2
x
in the equation
1
+
sin
x
=
2
cos
2
x
by
(
1
−
sin
2
x
)
⇒
2
(
1
−
sin
2
x
)
=
sin
x
+
1
or,
2
−
2
sin
2
x
=
sin
x
+
1
or,
0
=
2
sin
2
x
+
sin
x
+
1
−
2
or,
2
sin
2
x
+
sin
x
−
1
=
0
using the quadratic formula:
x
=
−
b
±
√
b
2
−
4
a
c
2
a
for quadratic equation
a
x
2
+
b
x
+
c
=
0
we have:
sin
x
=
−
1
±
√
1
2
−
4
⋅
2
⋅
(
−
1
)
2
⋅
2
or,
sin
x
=
−
1
±
√
1
+
8
4
or,
sin
x
=
−
1
±
√
9
4
or,
sin
x
=
−
1
±
3
4
or,
sin
x
=
−
1
+
3
4
,
−
1
−
3
4
or,
sin
x
=
1
2
,
−
1
Case I:
sin
x
=
1
2
for the condition:
0
≤
x
≤
2
π
we have:
x
=
π
6
or
5
π
6
to get positive value of
sin
x
Case II:
sin
x
=
−
1
we have:
x
=
3
π
2
to get negative value of
sin
x
Answer link
Объяснение: