(
6
1
)
4
=
6
4
1
=
1296
1
(−
2
1
)
6
=
2
6
1
=
64
1
(−3
3
1
)
3
=−(
3
10
)
3
=
=−
3
3
10
3
=−
27
1000
=−37
27
1
\begin{gathered}{4}^{3} + {3}^{5} = 64 + 243 = 307 \\ (0.6) ^{3} - (0.4) ^{3} = \\ = (\frac{3}{5} ) ^{3} - ( \frac{2}{5} )^{3} = \frac{27 - 8}{125} = \\ = \frac{19}{125}\end{gathered}
4
3
+3
5
=64+243=307
(0.6)
3
−(0.4)
3
=
=(
5
3
)
3
−(
5
2
)
3
=
125
27−8
=
=
125
19
\begin{gathered}0.12 \times {5}^{4} = 0.12 \times {5}^{2} \times {5}^{2} = \\ = 3 \times 25 = 75\end{gathered}
0.12×5
4
=0.12×5
2
×5
2
=
=3×25=75
4•(2-3m)-(6-m)-2•(3m+4)=
=8-12m-6+m-6m-8=
= -17m-6
(
6
1
)
4
=
6
4
1
=
1296
1
(−
2
1
)
6
=
2
6
1
=
64
1
(−3
3
1
)
3
=−(
3
10
)
3
=
=−
3
3
10
3
=−
27
1000
=−37
27
1
\begin{gathered}{4}^{3} + {3}^{5} = 64 + 243 = 307 \\ (0.6) ^{3} - (0.4) ^{3} = \\ = (\frac{3}{5} ) ^{3} - ( \frac{2}{5} )^{3} = \frac{27 - 8}{125} = \\ = \frac{19}{125}\end{gathered}
4
3
+3
5
=64+243=307
(0.6)
3
−(0.4)
3
=
=(
5
3
)
3
−(
5
2
)
3
=
125
27−8
=
=
125
19
\begin{gathered}0.12 \times {5}^{4} = 0.12 \times {5}^{2} \times {5}^{2} = \\ = 3 \times 25 = 75\end{gathered}
0.12×5
4
=0.12×5
2
×5
2
=
=3×25=75
4•(2-3m)-(6-m)-2•(3m+4)=
=8-12m-6+m-6m-8=
= -17m-6
Диагонали параллелограмма точкой пересечения делятся пополам.
AO=OC
P△BOC= BC+BO+OC = BC+BO+AO
P△AOB-P△BOC= AB+BO+AO -(BC+BO+AO) = AB-BC
Противоположные стороны параллелограмма равны.
P ABCD = 2(AB+BC)
{AB-BC=2
{2(AB+BC)=28 <=> AB+BC=14 <=> AB=14-BC
{14-BC-BC=2 <=> BC=6
{AB=14-6=8