e) \(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{68}}=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}{\dfrac{1}{4}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}=\dfrac{1}{3}:\dfrac{1}{4}=\dfrac{3}{4}\)

\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(\Rightarrow B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(\Rightarrow B=\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{10}{9.10}-\dfrac{9}{9.10}\)

\(\Rightarrow B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow B=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\\ B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\\ B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ B=\dfrac{1}{2}-\dfrac{1}{10}\\ B=\dfrac{5}{10}-\dfrac{1}{10}\\ B=\dfrac{4}{10}\\ B=\dfrac{2}{5}\)

a. Ta có:

b. Theo kết quả câu a,ta có:

a. Ta có:

b. Theo kết quả câu a,ta có:

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 = \(\dfrac{1}{4}.\dfrac{7}{3}.12\)

= \(\dfrac{1.7.12}{4.3}\)

= \(7\)
@Nguyễn Thành Đăng

B = \(\dfrac{3}{8}.56.\dfrac{25}{7}.\left(-4\right)\)

= \(-\dfrac{3.56.25.4}{8.7}\)

= -3.100
= -300
@Nguyễn Thành Đăng

a)\(\dfrac{6^2.6^3}{3^5}=\dfrac{2^2.3^2.2^3.3^3}{3^5}=2^5=32\)

b)\(\dfrac{25^2.4^2}{5^5\left(-2\right)^5}=\dfrac{5^2.5^2.2^2.2^2}{5^5.\left(-2\right)^5}=\dfrac{1}{-10}=-\dfrac{1}{10}\)

c)\(\dfrac{2^7.9^3}{8^2.3^6}=\dfrac{2^7.\left(3^2\right)^3}{\left(2^3\right)^2.3^6}=\dfrac{2^7.3^6}{2^6.3^6}=2\)

d)\(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\dfrac{3^3\left(2^3+2^2+1\right)}{-13}\)

\(=\dfrac{3^3.13}{-13}=-3^3=-27\)

\(C=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2^2\right)\\ C=\dfrac{6}{7}+\dfrac{5}{8}.\dfrac{1}{5}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\dfrac{-4}{1}\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{-3}{4}\\ C=\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{-42}{56}\\ C=\dfrac{97}{56}\)

\(A=15.\left(\dfrac{3}{5}-\dfrac{2}{3}\right)+1\)

\(A=15.\dfrac{-1}{15}+1\)

\(A=-1+1\)

\(A=0\)