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a) \(A=27\cdot36+73\cdot99+27\cdot14-49\cdot73\)
\(A=27\cdot\left(36+14\right)+73\cdot\left(99-49\right)\)
\(A=27\cdot50+73\cdot50\)
\(A=50\cdot\left(27+73\right)\)
\(A=50\cdot100\)
\(A=5000\)
b) \(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)
\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B-\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)
\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)
\(B=2^3\cdot5^3\)
\(B=10^3\)
\(B=1000\)
b) Đặt A = 1 + 2 + 22 + 23 + ..... + 299 + 2100
=> 2A = 2 + 22 + 23 + ..... + 2100 + 2101
=> 2A - A = 2101 - 1
=> A = 2101 - 1
c ) Đặt B = 5 + 53 + 55 + ..... + 595 + 597
=> 52B = 53 + 55 + ..... + 597 + 599
=> 25B - B = 599 - 5
=> 24B = 599 - 5
=> \(B=\frac{5^{99}-5}{24}\)
\(A=\left(4+\frac{1}{5}\right).\frac{19}{8}+\left(2+\frac{5}{8}\right).\frac{21}{5}\) =\(\frac{21}{5}.\frac{19}{8}+\frac{21}{8}.\frac{21}{5}\) =\(\frac{21}{5}.\left(\frac{19}{8}+\frac{21}{8}\right)\) = \(\frac{21}{5}.5\) =21 \(B=\frac{25}{2}.\left(3+\frac{2}{7}\right)-\frac{23}{7}.\left(5+\frac{1}{2}\right)\) =\(\frac{25}{2}.\frac{23}{7}-\frac{23}{7}.\frac{11}{2}\) =\(\frac{23}{7}.\left(\frac{25}{2}-\frac{11}{2}\right)\) =\(\frac{23}{7}.7=23\)
a=2^12.3^5-2^12.3^4/2^12.3^6+2^12.3^5 - 5^10.7^3-5^10.7^4/5^9.7^3+5^9.7^3.2^3
a=2^12.3^4.(3-1)/2^12.3^5.(3+1)-5^10.7^3.(1-7)/5^9.7^3.(1+8)
a=2/12-30/9
a=1/6-10/3=-19/6
a=
ính giá trị của các biểu thức sau:
A=827−(349+427)A=827−(349+427)
B=(1029+235)−629B=(1029+235)−629
Giải:
A=827−(349+427)A=827−(349+427)
=587−(319+307)=58−307−319=4−319=587−(319+307)=58−307−319=4−319
= 36−319=5936−319=59
B=(1029+235)−629B=(1029+235)−629
=1029−629+235=4+235=635
ính giá trị của các biểu thức sau:
A
=
8
2
7
−
(
3
4
9
+
4
2
7
)
A=827−(349+427)
B
=
(
10
2
9
+
2
3
5
)
−
6
2
9
B=(1029+235)−629
Giải:
A
=
8
2
7
−
(
3
4
9
+
4
2
7
)
A=827−(349+427)
=
58
7
−
(
31
9
+
30
7
)
=
58
−
30
7
−
31
9
=
4
−
31
9
=587−(319+307)=58−307−319=4−319
=
36
−
31
9
=
5
9
36−319=59
B
=
(
10
2
9
+
2
3
5
)
−
6
2
9
B=(1029+235)−629
=
10
2
9
−
6
2
9
+
2
3
5
=
4
+
2
3
5
=
6
3
5
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a, A = 27.36+73.99+27.14 - 49.73
= 27(36+14)+73(99 - 49)
= 27.50 + 73.50 = 50(27+73) = 50.100 = 5000
b, B = 4 5 . 10 . 5 6 + 25 5 . 2 8 : 2 8 . 5 4 + 5 7 . 2 5
Ta có: 4 5 = 2 2 5 = 2 2 . 5 = 2 10 ; 25 5 = 5 2 5 = 5 2 . 5 = 5 10
=> 4 5 . 10 . 5 6 + 25 5 . 2 8 = 2 10 . 2 . 5 . 5 6 + 5 10 . 2 8 = 2 11 . 5 7 + 5 10 . 2 8 = 2 8 . 2 3 . 5 7 + 5 7 . 5 3 . 2 8 = 2 8 . 5 7 . 2 3 + 5 3
Ta lại có: 2 8 . 5 4 + 5 7 . 2 5 = 2 5 . 2 3 . 5 4 + 5 4 . 5 3 . 2 5 = 2 5 . 5 4 . 2 3 + 5 3
Suy ra B = 2 8 . 5 7 . 2 3 + 5 3 :[ 2 5 . 5 4 . 2 3 + 5 3 ] = 2 8 . 5 7 : 2 5 . 5 4
= 2 8 : 2 5 . 5 7 : 5 4 = 2 3 . 5 3 = 2 . 5 3 = 10 3 = 1000
a, A = 27.36+73.99+27.14 - 49.73
= 27(36+14)+73(99 - 49)
= 27.50 + 73.50 = 50(27+73) = 50.100 = 5000
b, B = 4 5 . 10 . 5 6 + 25 5 . 2 8 : 2 8 . 5 4 + 5 7 . 2 5
Ta có: 4 5 = 2 2 5 = 2 2 . 5 = 2 10 ; 25 5 = 5 2 5 = 5 2 . 5 = 5 10
=> 4 5 . 10 . 5 6 + 25 5 . 2 8 = 2 10 . 2 . 5 . 5 6 + 5 10 . 2 8 = 2 11 . 5 7 + 5 10 . 2 8 = 2 8 . 2 3 . 5 7 + 5 7 . 5 3 . 2 8 = 2 8 . 5 7 . 2 3 + 5 3
Ta lại có: 2 8 . 5 4 + 5 7 . 2 5 = 2 5 . 2 3 . 5 4 + 5 4 . 5 3 . 2 5 = 2 5 . 5 4 . 2 3 + 5 3
Suy ra B = 2 8 . 5 7 . 2 3 + 5 3 :[ 2 5 . 5 4 . 2 3 + 5 3 ] = 2 8 . 5 7 : 2 5 . 5 4
= 2 8 : 2 5 . 5 7 : 5 4 = 2 3 . 5 3 = 2 . 5 3 = 10 3 = 1000