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câu a
\(A=\frac{33.10^3}{2^3.5.10^3+7000}=\frac{33.10^3}{2^3.5.10^3+7.10^3}=\frac{33.10^3}{10^3\left(2^3.5+7\right)}=\frac{33.10^3}{10^3.47}=\frac{33}{47}\)
\(B=\frac{3774}{5217}=\frac{34.111}{47.111}=\frac{34}{47}\)
\(\Rightarrow\frac{33}{47}< \frac{34}{47}\)
=> A<B
e. \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}=-\frac{3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)=-\frac{3}{5}\cdot2=-\frac{6}{5}\)
f. \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{5}=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{4}{5}=\frac{1}{3}\cdot2-\frac{4}{5}=\frac{2}{3}-\frac{4}{5}=-\frac{2}{15}\)
g. \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{15}{19}+\frac{5}{7}=-\frac{3}{7}\left(\frac{4}{19}+\frac{15}{19}\right)+\frac{5}{7}=-\frac{3}{7}\cdot1+\frac{5}{7}=-\frac{3}{7}+\frac{5}{7}=\frac{2}{7}\)
h. \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
\(A=5+5^2+5^3+5^4+...+5^{2004}\)
\(5A=5^2+5^3+5^4+5^5+...+5^{2005}\)
\(5A-A=\left(5^2+5^3+5^4+5^5+...+5^{2005}\right)-\left(5+5^2+5^3+5^4+...+5^{2004}\right)\)
\(4A=5^{2005}-5\)
\(A=\dfrac{5^{2005}-5}{4}\)
\(B=7^1+7^2+7^3+....+7^{2015}\)
\(7B=7^2+7^3+7^4+....+7^{2016}\)
\(7B-B=\left(7^2+7^3+7^4+...+7^{2016}\right)-\left(7+7^2+7^3+....+7^{2015}\right)\)
\(6B=7^{2016}-7\)
\(B=\dfrac{7^{2016}-7}{6}\)
\(C=4^5+4^6+4^7+...+4^{2016}\)
\(4C=4^6+4^7+4^8+...+4^{2017}\)
\(4C-C=\left(4^6+4^7+4^8+...+4^{2017}\right)-\left(4^5+4^6+4^7+...+4^{2016}\right)\)
\(3C=4^{2017}-4^5\)
\(C=\dfrac{4^{2017}-4^5}{3}\)
A = 5 + 52 + 53 + 54 + ... + 52004
5A = 52 + 53 + 54 + 55 + ... + 52005
5A - A = 52005 - 5
4A = 52005 - 5
A = (52005 - 5) : 4
B = 71 + 72 + 73 + ... + 72015
7B = 72 + 73 + 74 + ... + 72016
7B - B = 72016 - 7
6B = 72016 - 7
B = (72016 - 7) : 6
C = 45 + 46 + 47 + ... + 42016
4C = 46 + 47 + 48 + ... + 42017
4C - C = 42017 - 45
3C = 42017 - 45
C = (42017 - 45) : 3
Bài 1:
a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)
\(=\frac{9}{15}+\frac{4}{15}\)
\(=\frac{13}{15}\)
b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)
\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)
c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)
\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)
\(=\frac{-7}{8}:\frac{-7}{4}\)
\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)
e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)
f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)
\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)
\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)
\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)
\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)
h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)
\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)
\(=\frac{5}{9}\)
a ) Ta có : 4(x - 5) - 3(x + 7) = -19
<=> 4x - 20 - 3x - 21 = -19
=> x - 41 = -19
=> x = -19 + 41
=> x = 22
b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5
<=> 7x - 21 - 15 + 5x = 11x - 5
<=> 12x - 36 = 11x - 5
=> 12x - 11x = -5 + 36
=> x = 31
a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)
\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)
\(=4-1\dfrac{3}{4}\)
\(=3\dfrac{3}{4}\)
b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)
\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)
\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)
\(=4-2\dfrac{3}{7}\)
\(=2\dfrac{3}{7}\)
Ta có:
57 : 53 = 54
57 : 54 = 53
57:53=1075471698
57:54=1055555556